Evaluate $sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)}$












5












$begingroup$



Evaluate
$$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)}$$




So in a previous part of the question I calculated that
$$sum_{r=1}^{n} frac{2-r}{r(r+1)(r+2)} = sum_{r=1}^{n}left( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=frac{n}{(n+1)(n+2)}$$



So my question is then why does $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)} = -frac{1}{6}$$



As I though that it would be $frac{1}{2}$.






summation







share|cite|improve this question









$endgroup$

















    5












    $begingroup$



    Evaluate
    $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)}$$




    So in a previous part of the question I calculated that
    $$sum_{r=1}^{n} frac{2-r}{r(r+1)(r+2)} = sum_{r=1}^{n}left( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=frac{n}{(n+1)(n+2)}$$



    So my question is then why does $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)} = -frac{1}{6}$$



    As I though that it would be $frac{1}{2}$.






    summation







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$



      Evaluate
      $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)}$$




      So in a previous part of the question I calculated that
      $$sum_{r=1}^{n} frac{2-r}{r(r+1)(r+2)} = sum_{r=1}^{n}left( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=frac{n}{(n+1)(n+2)}$$



      So my question is then why does $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)} = -frac{1}{6}$$



      As I though that it would be $frac{1}{2}$.






      summation







      share|cite|improve this question









      $endgroup$





      Evaluate
      $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)}$$




      So in a previous part of the question I calculated that
      $$sum_{r=1}^{n} frac{2-r}{r(r+1)(r+2)} = sum_{r=1}^{n}left( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=frac{n}{(n+1)(n+2)}$$



      So my question is then why does $$sum_{r=2}^{infty} frac{2-r}{r(r+1)(r+2)} = -frac{1}{6}$$



      As I though that it would be $frac{1}{2}$.






      summation




      summation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 15:47









      H.LinkhornH.Linkhorn

      537313




      537313






















          6 Answers
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          $begingroup$

          Note thatbegin{align}frac{2-r}{r(r+1)(r+2)}&=frac1r-frac3{r+1}+frac2{r+2}\&=frac1r-frac1{r+1}-frac2{r+1}+frac2{r+2}.end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$frac12-frac23=-frac16.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) was a minute late!
            $endgroup$
            – TheSimpliFire
            Dec 31 '18 at 15:54



















          5












          $begingroup$

          Note that begin{align}sum_{r=2}^inftyleft( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)&=sum_{r=2}^inftyleft(frac1r-frac1{r+1}right)-2sum_{r=2}^inftyleft(frac1{r+1}-frac1{r+2}right)\&=left(frac12-frac13+frac13-frac14+cdotsright)-2left(frac13+frac14-frac14+frac15-cdotsright)\&=frac12-frac23=-frac16end{align}






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            (+1) Believe me: I know what it feels to be a few seconds too late.
            $endgroup$
            – José Carlos Santos
            Dec 31 '18 at 15:55



















          1












          $begingroup$

          Actually your finite sum is equal to $$sum_{r=2}^nfrac{2-r}{r(r+1)(r+2)}=frac{2}{n+2}-frac{1}{n+1}-frac{1}{6}$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let $dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=dfrac{am+b}{m(m+1)}$



            $implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$



            Set $2b=2iff b=1,ar=-riff a=-1$



            $$impliessum_{r=1}^ndfrac{2-r}{r(r+1)(r+2)}=sum_{r=1}^nleft(f(r)-f(r+1)right)=f(1)-f(n+1)$$



            Set $ntoinfty$ and show that $ntoinfty f(n)=0$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              If $H_{n}$ is the $n$-th harmonic number, we have $$sum_{r=2}^{n}left(frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=left(H_{n}-1right)-3left(H_{n}-1-frac{1}{2}right)+2left(H_{n}-1-frac{1}{2}-frac{1}{3}right)$$ $$=-1+frac{3}{2}-frac{2}{3}=-frac{1}{6}.$$ Now take the limit as $nrightarrow+infty$.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                By definition,



                $$ sum_{r=1}^infty a_r = lim_{ntoinfty}sum_{r=1}^n a_r .$$



                In your case the above limit is zero. Then:



                $$ sum_{r=2}^infty a_r = sum_{r=1}^infty a_r - a_1 = -a_1 . $$



                This reasoning is what @HenriLee was referring to in his answer.






                share|cite|improve this answer









                $endgroup$














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                  6 Answers
                  6






                  active

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                  6 Answers
                  6






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  7












                  $begingroup$

                  Note thatbegin{align}frac{2-r}{r(r+1)(r+2)}&=frac1r-frac3{r+1}+frac2{r+2}\&=frac1r-frac1{r+1}-frac2{r+1}+frac2{r+2}.end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$frac12-frac23=-frac16.$$






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    (+1) was a minute late!
                    $endgroup$
                    – TheSimpliFire
                    Dec 31 '18 at 15:54
















                  7












                  $begingroup$

                  Note thatbegin{align}frac{2-r}{r(r+1)(r+2)}&=frac1r-frac3{r+1}+frac2{r+2}\&=frac1r-frac1{r+1}-frac2{r+1}+frac2{r+2}.end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$frac12-frac23=-frac16.$$






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    (+1) was a minute late!
                    $endgroup$
                    – TheSimpliFire
                    Dec 31 '18 at 15:54














                  7












                  7








                  7





                  $begingroup$

                  Note thatbegin{align}frac{2-r}{r(r+1)(r+2)}&=frac1r-frac3{r+1}+frac2{r+2}\&=frac1r-frac1{r+1}-frac2{r+1}+frac2{r+2}.end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$frac12-frac23=-frac16.$$






                  share|cite|improve this answer









                  $endgroup$



                  Note thatbegin{align}frac{2-r}{r(r+1)(r+2)}&=frac1r-frac3{r+1}+frac2{r+2}\&=frac1r-frac1{r+1}-frac2{r+1}+frac2{r+2}.end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$frac12-frac23=-frac16.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 15:52









                  José Carlos SantosJosé Carlos Santos

                  178k24139251




                  178k24139251












                  • $begingroup$
                    (+1) was a minute late!
                    $endgroup$
                    – TheSimpliFire
                    Dec 31 '18 at 15:54


















                  • $begingroup$
                    (+1) was a minute late!
                    $endgroup$
                    – TheSimpliFire
                    Dec 31 '18 at 15:54
















                  $begingroup$
                  (+1) was a minute late!
                  $endgroup$
                  – TheSimpliFire
                  Dec 31 '18 at 15:54




                  $begingroup$
                  (+1) was a minute late!
                  $endgroup$
                  – TheSimpliFire
                  Dec 31 '18 at 15:54











                  5












                  $begingroup$

                  Note that begin{align}sum_{r=2}^inftyleft( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)&=sum_{r=2}^inftyleft(frac1r-frac1{r+1}right)-2sum_{r=2}^inftyleft(frac1{r+1}-frac1{r+2}right)\&=left(frac12-frac13+frac13-frac14+cdotsright)-2left(frac13+frac14-frac14+frac15-cdotsright)\&=frac12-frac23=-frac16end{align}






                  share|cite|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    (+1) Believe me: I know what it feels to be a few seconds too late.
                    $endgroup$
                    – José Carlos Santos
                    Dec 31 '18 at 15:55
















                  5












                  $begingroup$

                  Note that begin{align}sum_{r=2}^inftyleft( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)&=sum_{r=2}^inftyleft(frac1r-frac1{r+1}right)-2sum_{r=2}^inftyleft(frac1{r+1}-frac1{r+2}right)\&=left(frac12-frac13+frac13-frac14+cdotsright)-2left(frac13+frac14-frac14+frac15-cdotsright)\&=frac12-frac23=-frac16end{align}






                  share|cite|improve this answer









                  $endgroup$









                  • 1




                    $begingroup$
                    (+1) Believe me: I know what it feels to be a few seconds too late.
                    $endgroup$
                    – José Carlos Santos
                    Dec 31 '18 at 15:55














                  5












                  5








                  5





                  $begingroup$

                  Note that begin{align}sum_{r=2}^inftyleft( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)&=sum_{r=2}^inftyleft(frac1r-frac1{r+1}right)-2sum_{r=2}^inftyleft(frac1{r+1}-frac1{r+2}right)\&=left(frac12-frac13+frac13-frac14+cdotsright)-2left(frac13+frac14-frac14+frac15-cdotsright)\&=frac12-frac23=-frac16end{align}






                  share|cite|improve this answer









                  $endgroup$



                  Note that begin{align}sum_{r=2}^inftyleft( frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)&=sum_{r=2}^inftyleft(frac1r-frac1{r+1}right)-2sum_{r=2}^inftyleft(frac1{r+1}-frac1{r+2}right)\&=left(frac12-frac13+frac13-frac14+cdotsright)-2left(frac13+frac14-frac14+frac15-cdotsright)\&=frac12-frac23=-frac16end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 31 '18 at 15:53









                  TheSimpliFireTheSimpliFire

                  13.6k62765




                  13.6k62765








                  • 1




                    $begingroup$
                    (+1) Believe me: I know what it feels to be a few seconds too late.
                    $endgroup$
                    – José Carlos Santos
                    Dec 31 '18 at 15:55














                  • 1




                    $begingroup$
                    (+1) Believe me: I know what it feels to be a few seconds too late.
                    $endgroup$
                    – José Carlos Santos
                    Dec 31 '18 at 15:55








                  1




                  1




                  $begingroup$
                  (+1) Believe me: I know what it feels to be a few seconds too late.
                  $endgroup$
                  – José Carlos Santos
                  Dec 31 '18 at 15:55




                  $begingroup$
                  (+1) Believe me: I know what it feels to be a few seconds too late.
                  $endgroup$
                  – José Carlos Santos
                  Dec 31 '18 at 15:55











                  1












                  $begingroup$

                  Actually your finite sum is equal to $$sum_{r=2}^nfrac{2-r}{r(r+1)(r+2)}=frac{2}{n+2}-frac{1}{n+1}-frac{1}{6}$$






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Actually your finite sum is equal to $$sum_{r=2}^nfrac{2-r}{r(r+1)(r+2)}=frac{2}{n+2}-frac{1}{n+1}-frac{1}{6}$$






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Actually your finite sum is equal to $$sum_{r=2}^nfrac{2-r}{r(r+1)(r+2)}=frac{2}{n+2}-frac{1}{n+1}-frac{1}{6}$$






                      share|cite|improve this answer









                      $endgroup$



                      Actually your finite sum is equal to $$sum_{r=2}^nfrac{2-r}{r(r+1)(r+2)}=frac{2}{n+2}-frac{1}{n+1}-frac{1}{6}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 31 '18 at 15:54









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      79.9k42867




                      79.9k42867























                          1












                          $begingroup$

                          Let $dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=dfrac{am+b}{m(m+1)}$



                          $implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$



                          Set $2b=2iff b=1,ar=-riff a=-1$



                          $$impliessum_{r=1}^ndfrac{2-r}{r(r+1)(r+2)}=sum_{r=1}^nleft(f(r)-f(r+1)right)=f(1)-f(n+1)$$



                          Set $ntoinfty$ and show that $ntoinfty f(n)=0$






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            Let $dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=dfrac{am+b}{m(m+1)}$



                            $implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$



                            Set $2b=2iff b=1,ar=-riff a=-1$



                            $$impliessum_{r=1}^ndfrac{2-r}{r(r+1)(r+2)}=sum_{r=1}^nleft(f(r)-f(r+1)right)=f(1)-f(n+1)$$



                            Set $ntoinfty$ and show that $ntoinfty f(n)=0$






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Let $dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=dfrac{am+b}{m(m+1)}$



                              $implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$



                              Set $2b=2iff b=1,ar=-riff a=-1$



                              $$impliessum_{r=1}^ndfrac{2-r}{r(r+1)(r+2)}=sum_{r=1}^nleft(f(r)-f(r+1)right)=f(1)-f(n+1)$$



                              Set $ntoinfty$ and show that $ntoinfty f(n)=0$






                              share|cite|improve this answer









                              $endgroup$



                              Let $dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=dfrac{am+b}{m(m+1)}$



                              $implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$



                              Set $2b=2iff b=1,ar=-riff a=-1$



                              $$impliessum_{r=1}^ndfrac{2-r}{r(r+1)(r+2)}=sum_{r=1}^nleft(f(r)-f(r+1)right)=f(1)-f(n+1)$$



                              Set $ntoinfty$ and show that $ntoinfty f(n)=0$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 31 '18 at 16:07









                              lab bhattacharjeelab bhattacharjee

                              229k15159280




                              229k15159280























                                  1












                                  $begingroup$

                                  If $H_{n}$ is the $n$-th harmonic number, we have $$sum_{r=2}^{n}left(frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=left(H_{n}-1right)-3left(H_{n}-1-frac{1}{2}right)+2left(H_{n}-1-frac{1}{2}-frac{1}{3}right)$$ $$=-1+frac{3}{2}-frac{2}{3}=-frac{1}{6}.$$ Now take the limit as $nrightarrow+infty$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    If $H_{n}$ is the $n$-th harmonic number, we have $$sum_{r=2}^{n}left(frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=left(H_{n}-1right)-3left(H_{n}-1-frac{1}{2}right)+2left(H_{n}-1-frac{1}{2}-frac{1}{3}right)$$ $$=-1+frac{3}{2}-frac{2}{3}=-frac{1}{6}.$$ Now take the limit as $nrightarrow+infty$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      If $H_{n}$ is the $n$-th harmonic number, we have $$sum_{r=2}^{n}left(frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=left(H_{n}-1right)-3left(H_{n}-1-frac{1}{2}right)+2left(H_{n}-1-frac{1}{2}-frac{1}{3}right)$$ $$=-1+frac{3}{2}-frac{2}{3}=-frac{1}{6}.$$ Now take the limit as $nrightarrow+infty$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      If $H_{n}$ is the $n$-th harmonic number, we have $$sum_{r=2}^{n}left(frac{1}{r}-frac{3}{r+1}+frac{2}{r+2}right)=left(H_{n}-1right)-3left(H_{n}-1-frac{1}{2}right)+2left(H_{n}-1-frac{1}{2}-frac{1}{3}right)$$ $$=-1+frac{3}{2}-frac{2}{3}=-frac{1}{6}.$$ Now take the limit as $nrightarrow+infty$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 31 '18 at 16:08









                                      Marco CantariniMarco Cantarini

                                      28.8k23373




                                      28.8k23373























                                          0












                                          $begingroup$

                                          By definition,



                                          $$ sum_{r=1}^infty a_r = lim_{ntoinfty}sum_{r=1}^n a_r .$$



                                          In your case the above limit is zero. Then:



                                          $$ sum_{r=2}^infty a_r = sum_{r=1}^infty a_r - a_1 = -a_1 . $$



                                          This reasoning is what @HenriLee was referring to in his answer.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            By definition,



                                            $$ sum_{r=1}^infty a_r = lim_{ntoinfty}sum_{r=1}^n a_r .$$



                                            In your case the above limit is zero. Then:



                                            $$ sum_{r=2}^infty a_r = sum_{r=1}^infty a_r - a_1 = -a_1 . $$



                                            This reasoning is what @HenriLee was referring to in his answer.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              By definition,



                                              $$ sum_{r=1}^infty a_r = lim_{ntoinfty}sum_{r=1}^n a_r .$$



                                              In your case the above limit is zero. Then:



                                              $$ sum_{r=2}^infty a_r = sum_{r=1}^infty a_r - a_1 = -a_1 . $$



                                              This reasoning is what @HenriLee was referring to in his answer.






                                              share|cite|improve this answer









                                              $endgroup$



                                              By definition,



                                              $$ sum_{r=1}^infty a_r = lim_{ntoinfty}sum_{r=1}^n a_r .$$



                                              In your case the above limit is zero. Then:



                                              $$ sum_{r=2}^infty a_r = sum_{r=1}^infty a_r - a_1 = -a_1 . $$



                                              This reasoning is what @HenriLee was referring to in his answer.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 31 '18 at 15:57









                                              Dog_69Dog_69

                                              6361523




                                              6361523






























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                                                  Bundesstraße 106

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                                                  Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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                                                  Ida-Boy-Ed-Garten

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                                                  Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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                                                  Verónica Boquete

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                                                  Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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