Determine whether $sum_{n=1}^{infty}frac{n2^n}{n+3^n}$ converges.












2












$begingroup$


L'Hôspital is not allowed, we are allowed to use algebraic methods.



First I have tried the root-criteria:



$$limsup_{nrightarrow infty}sqrt[n]frac{n2^n}{n+3^n}iff lim_{nrightarrow infty}frac{2}{sqrt[n]{n+3^n}},.$$ I don't know how to proceed further.



Then I have tried for a long time the quotient criteria, i.e.,



$$limsup_{nrightarrow infty}frac{frac{(n+1)2^{n+1}}{n+1+3^{n+1}}}{frac{n2^n}{n+3^n}}iff 2lim_{nrightarrow infty}frac{n+3^n}{n+1+3^{n+1}},.$$



Last I have tried Discrete Comparison Test. My idea comparing



$$frac{n2^n}{n+3^n}text{ vs }frac{1}{n},,text{ i.e. },frac{n^22^n}{n+3^n}text{ vs }1,.$$



I have seen the plots on WolframAlpha and I have seen that $n^22^n$ converges faster then $n+3^n$ therefore $frac{n2^n}{n+3^n}>frac{1}{n}$ for almost all $n$ therefore it diverges, but I am not sure whether one can argue like that. And how to prove that $n^22^n$ converges faster then $n+3^n$ formally.



Thanks for the support!






calculus sequences-and-series convergence limits-without-lhopital







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$endgroup$








  • 1




    $begingroup$
    It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
    $endgroup$
    – K. 622
    Dec 15 '18 at 0:34
















2












$begingroup$


L'Hôspital is not allowed, we are allowed to use algebraic methods.



First I have tried the root-criteria:



$$limsup_{nrightarrow infty}sqrt[n]frac{n2^n}{n+3^n}iff lim_{nrightarrow infty}frac{2}{sqrt[n]{n+3^n}},.$$ I don't know how to proceed further.



Then I have tried for a long time the quotient criteria, i.e.,



$$limsup_{nrightarrow infty}frac{frac{(n+1)2^{n+1}}{n+1+3^{n+1}}}{frac{n2^n}{n+3^n}}iff 2lim_{nrightarrow infty}frac{n+3^n}{n+1+3^{n+1}},.$$



Last I have tried Discrete Comparison Test. My idea comparing



$$frac{n2^n}{n+3^n}text{ vs }frac{1}{n},,text{ i.e. },frac{n^22^n}{n+3^n}text{ vs }1,.$$



I have seen the plots on WolframAlpha and I have seen that $n^22^n$ converges faster then $n+3^n$ therefore $frac{n2^n}{n+3^n}>frac{1}{n}$ for almost all $n$ therefore it diverges, but I am not sure whether one can argue like that. And how to prove that $n^22^n$ converges faster then $n+3^n$ formally.



Thanks for the support!






calculus sequences-and-series convergence limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
    $endgroup$
    – K. 622
    Dec 15 '18 at 0:34














2












2








2





$begingroup$


L'Hôspital is not allowed, we are allowed to use algebraic methods.



First I have tried the root-criteria:



$$limsup_{nrightarrow infty}sqrt[n]frac{n2^n}{n+3^n}iff lim_{nrightarrow infty}frac{2}{sqrt[n]{n+3^n}},.$$ I don't know how to proceed further.



Then I have tried for a long time the quotient criteria, i.e.,



$$limsup_{nrightarrow infty}frac{frac{(n+1)2^{n+1}}{n+1+3^{n+1}}}{frac{n2^n}{n+3^n}}iff 2lim_{nrightarrow infty}frac{n+3^n}{n+1+3^{n+1}},.$$



Last I have tried Discrete Comparison Test. My idea comparing



$$frac{n2^n}{n+3^n}text{ vs }frac{1}{n},,text{ i.e. },frac{n^22^n}{n+3^n}text{ vs }1,.$$



I have seen the plots on WolframAlpha and I have seen that $n^22^n$ converges faster then $n+3^n$ therefore $frac{n2^n}{n+3^n}>frac{1}{n}$ for almost all $n$ therefore it diverges, but I am not sure whether one can argue like that. And how to prove that $n^22^n$ converges faster then $n+3^n$ formally.



Thanks for the support!






calculus sequences-and-series convergence limits-without-lhopital







share|cite|improve this question











$endgroup$




L'Hôspital is not allowed, we are allowed to use algebraic methods.



First I have tried the root-criteria:



$$limsup_{nrightarrow infty}sqrt[n]frac{n2^n}{n+3^n}iff lim_{nrightarrow infty}frac{2}{sqrt[n]{n+3^n}},.$$ I don't know how to proceed further.



Then I have tried for a long time the quotient criteria, i.e.,



$$limsup_{nrightarrow infty}frac{frac{(n+1)2^{n+1}}{n+1+3^{n+1}}}{frac{n2^n}{n+3^n}}iff 2lim_{nrightarrow infty}frac{n+3^n}{n+1+3^{n+1}},.$$



Last I have tried Discrete Comparison Test. My idea comparing



$$frac{n2^n}{n+3^n}text{ vs }frac{1}{n},,text{ i.e. },frac{n^22^n}{n+3^n}text{ vs }1,.$$



I have seen the plots on WolframAlpha and I have seen that $n^22^n$ converges faster then $n+3^n$ therefore $frac{n2^n}{n+3^n}>frac{1}{n}$ for almost all $n$ therefore it diverges, but I am not sure whether one can argue like that. And how to prove that $n^22^n$ converges faster then $n+3^n$ formally.



Thanks for the support!






calculus sequences-and-series convergence limits-without-lhopital




calculus sequences-and-series convergence limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 20:58









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 20:43









RM777RM777

36912




36912








  • 1




    $begingroup$
    It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
    $endgroup$
    – K. 622
    Dec 15 '18 at 0:34














  • 1




    $begingroup$
    It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
    $endgroup$
    – K. 622
    Dec 15 '18 at 0:34








1




1




$begingroup$
It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
$endgroup$
– K. 622
Dec 15 '18 at 0:34




$begingroup$
It's l'Hôpital (or l'Hospital) by the way, the circumflex represents a lost consonant, in this case 's'.
$endgroup$
– K. 622
Dec 15 '18 at 0:34










3 Answers
3






active

oldest

votes


















4












$begingroup$

HINT:



Note that we have



$$0le frac{n2^n}{n+3^n}le nleft(frac23right)^n$$



Can you finish now?




If you want to apply the root test directly, then



$$begin{align}
lim_{nto infty}sqrt[n]{frac{n2^n}{n+3^n}}&=frac23lim_{nto infty}left(frac{n^{1/n}}{left(1+frac n{3^n}right)^{1/n}}right)\\
&=frac23lim_{nto infty}left( n^{1/n}e^{-frac1n logleft(1+frac n{3^n}right)}right)
end{align}$$



Can you evaluate the limit $lim_{nto infty}frac{logleft(1+frac n{3^n}right)}{n}$?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you $n(frac{2}{3})^n$ is a convergent majorant.
    $endgroup$
    – RM777
    Dec 15 '18 at 18:19










  • $begingroup$
    You're welcome. My pleasure.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 19:02



















2












$begingroup$

We may use the ratio test.



Consider $$|dfrac{a_{n+1}}{a_n}|=left|dfrac{dfrac{n2^{n+1}}{n+1+3^{n+1}}}{dfrac{n2^n}{n+3^n}}right|=left|dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}right|$$



Taking the limit as $ntoinfty$, we see it approaches $2/3 lt 1$.



Therefore by ratio test, your series converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 4:49





















0












$begingroup$

As an alternative by ratio test



$$frac{(n+1)2^{n+1}}{n+1+3^{n+1}}frac{3^n}{n2^n}=frac{2(n+1)}{n}frac{n+3^n}{n+1+3^{n+1}}to frac 23$$



and therefore the series converges.






share|cite|improve this answer











$endgroup$













    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    HINT:



    Note that we have



    $$0le frac{n2^n}{n+3^n}le nleft(frac23right)^n$$



    Can you finish now?




    If you want to apply the root test directly, then



    $$begin{align}
    lim_{nto infty}sqrt[n]{frac{n2^n}{n+3^n}}&=frac23lim_{nto infty}left(frac{n^{1/n}}{left(1+frac n{3^n}right)^{1/n}}right)\\
    &=frac23lim_{nto infty}left( n^{1/n}e^{-frac1n logleft(1+frac n{3^n}right)}right)
    end{align}$$



    Can you evaluate the limit $lim_{nto infty}frac{logleft(1+frac n{3^n}right)}{n}$?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you $n(frac{2}{3})^n$ is a convergent majorant.
      $endgroup$
      – RM777
      Dec 15 '18 at 18:19










    • $begingroup$
      You're welcome. My pleasure.
      $endgroup$
      – Mark Viola
      Dec 15 '18 at 19:02
















    4












    $begingroup$

    HINT:



    Note that we have



    $$0le frac{n2^n}{n+3^n}le nleft(frac23right)^n$$



    Can you finish now?




    If you want to apply the root test directly, then



    $$begin{align}
    lim_{nto infty}sqrt[n]{frac{n2^n}{n+3^n}}&=frac23lim_{nto infty}left(frac{n^{1/n}}{left(1+frac n{3^n}right)^{1/n}}right)\\
    &=frac23lim_{nto infty}left( n^{1/n}e^{-frac1n logleft(1+frac n{3^n}right)}right)
    end{align}$$



    Can you evaluate the limit $lim_{nto infty}frac{logleft(1+frac n{3^n}right)}{n}$?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you $n(frac{2}{3})^n$ is a convergent majorant.
      $endgroup$
      – RM777
      Dec 15 '18 at 18:19










    • $begingroup$
      You're welcome. My pleasure.
      $endgroup$
      – Mark Viola
      Dec 15 '18 at 19:02














    4












    4








    4





    $begingroup$

    HINT:



    Note that we have



    $$0le frac{n2^n}{n+3^n}le nleft(frac23right)^n$$



    Can you finish now?




    If you want to apply the root test directly, then



    $$begin{align}
    lim_{nto infty}sqrt[n]{frac{n2^n}{n+3^n}}&=frac23lim_{nto infty}left(frac{n^{1/n}}{left(1+frac n{3^n}right)^{1/n}}right)\\
    &=frac23lim_{nto infty}left( n^{1/n}e^{-frac1n logleft(1+frac n{3^n}right)}right)
    end{align}$$



    Can you evaluate the limit $lim_{nto infty}frac{logleft(1+frac n{3^n}right)}{n}$?






    share|cite|improve this answer











    $endgroup$



    HINT:



    Note that we have



    $$0le frac{n2^n}{n+3^n}le nleft(frac23right)^n$$



    Can you finish now?




    If you want to apply the root test directly, then



    $$begin{align}
    lim_{nto infty}sqrt[n]{frac{n2^n}{n+3^n}}&=frac23lim_{nto infty}left(frac{n^{1/n}}{left(1+frac n{3^n}right)^{1/n}}right)\\
    &=frac23lim_{nto infty}left( n^{1/n}e^{-frac1n logleft(1+frac n{3^n}right)}right)
    end{align}$$



    Can you evaluate the limit $lim_{nto infty}frac{logleft(1+frac n{3^n}right)}{n}$?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 14 '18 at 20:51

























    answered Dec 14 '18 at 20:44









    Mark ViolaMark Viola

    133k1277175




    133k1277175












    • $begingroup$
      Thank you $n(frac{2}{3})^n$ is a convergent majorant.
      $endgroup$
      – RM777
      Dec 15 '18 at 18:19










    • $begingroup$
      You're welcome. My pleasure.
      $endgroup$
      – Mark Viola
      Dec 15 '18 at 19:02


















    • $begingroup$
      Thank you $n(frac{2}{3})^n$ is a convergent majorant.
      $endgroup$
      – RM777
      Dec 15 '18 at 18:19










    • $begingroup$
      You're welcome. My pleasure.
      $endgroup$
      – Mark Viola
      Dec 15 '18 at 19:02
















    $begingroup$
    Thank you $n(frac{2}{3})^n$ is a convergent majorant.
    $endgroup$
    – RM777
    Dec 15 '18 at 18:19




    $begingroup$
    Thank you $n(frac{2}{3})^n$ is a convergent majorant.
    $endgroup$
    – RM777
    Dec 15 '18 at 18:19












    $begingroup$
    You're welcome. My pleasure.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 19:02




    $begingroup$
    You're welcome. My pleasure.
    $endgroup$
    – Mark Viola
    Dec 15 '18 at 19:02











    2












    $begingroup$

    We may use the ratio test.



    Consider $$|dfrac{a_{n+1}}{a_n}|=left|dfrac{dfrac{n2^{n+1}}{n+1+3^{n+1}}}{dfrac{n2^n}{n+3^n}}right|=left|dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}right|$$



    Taking the limit as $ntoinfty$, we see it approaches $2/3 lt 1$.



    Therefore by ratio test, your series converges.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 4:49


















    2












    $begingroup$

    We may use the ratio test.



    Consider $$|dfrac{a_{n+1}}{a_n}|=left|dfrac{dfrac{n2^{n+1}}{n+1+3^{n+1}}}{dfrac{n2^n}{n+3^n}}right|=left|dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}right|$$



    Taking the limit as $ntoinfty$, we see it approaches $2/3 lt 1$.



    Therefore by ratio test, your series converges.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 4:49
















    2












    2








    2





    $begingroup$

    We may use the ratio test.



    Consider $$|dfrac{a_{n+1}}{a_n}|=left|dfrac{dfrac{n2^{n+1}}{n+1+3^{n+1}}}{dfrac{n2^n}{n+3^n}}right|=left|dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}right|$$



    Taking the limit as $ntoinfty$, we see it approaches $2/3 lt 1$.



    Therefore by ratio test, your series converges.






    share|cite|improve this answer









    $endgroup$



    We may use the ratio test.



    Consider $$|dfrac{a_{n+1}}{a_n}|=left|dfrac{dfrac{n2^{n+1}}{n+1+3^{n+1}}}{dfrac{n2^n}{n+3^n}}right|=left|dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}right|$$



    Taking the limit as $ntoinfty$, we see it approaches $2/3 lt 1$.



    Therefore by ratio test, your series converges.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 14 '18 at 21:01









    K Split XK Split X

    4,30421233




    4,30421233












    • $begingroup$
      We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 4:49




















    • $begingroup$
      We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
      $endgroup$
      – DanielWainfleet
      Dec 15 '18 at 4:49


















    $begingroup$
    We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 4:49






    $begingroup$
    We can also apply the ratio test to $n2^n/3^n,$ and then use the comparison test, as $0<a_n<n2^n/3^n$
    $endgroup$
    – DanielWainfleet
    Dec 15 '18 at 4:49













    0












    $begingroup$

    As an alternative by ratio test



    $$frac{(n+1)2^{n+1}}{n+1+3^{n+1}}frac{3^n}{n2^n}=frac{2(n+1)}{n}frac{n+3^n}{n+1+3^{n+1}}to frac 23$$



    and therefore the series converges.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      As an alternative by ratio test



      $$frac{(n+1)2^{n+1}}{n+1+3^{n+1}}frac{3^n}{n2^n}=frac{2(n+1)}{n}frac{n+3^n}{n+1+3^{n+1}}to frac 23$$



      and therefore the series converges.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        As an alternative by ratio test



        $$frac{(n+1)2^{n+1}}{n+1+3^{n+1}}frac{3^n}{n2^n}=frac{2(n+1)}{n}frac{n+3^n}{n+1+3^{n+1}}to frac 23$$



        and therefore the series converges.






        share|cite|improve this answer











        $endgroup$



        As an alternative by ratio test



        $$frac{(n+1)2^{n+1}}{n+1+3^{n+1}}frac{3^n}{n2^n}=frac{2(n+1)}{n}frac{n+3^n}{n+1+3^{n+1}}to frac 23$$



        and therefore the series converges.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 15 '18 at 0:06

























        answered Dec 14 '18 at 21:01









        gimusigimusi

        92.9k84494




        92.9k84494






























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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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            Residenzschloss Arolsen

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            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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