Examining the following Series for convergence with proof












-1














Good day dear community,



I have to examine the following series for covergence and proof it.



(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$



(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$



My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$



My approach to (ii)



$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$



$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$



$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.



My approach to (iii)



And here I have no idea, whatsoever.






limits proof-verification convergence







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  • For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
    – Nodt Greenish
    Nov 25 at 14:27












  • For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
    – xbh
    Nov 25 at 14:27








  • 1




    $(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
    – Galc127
    Nov 25 at 14:29










  • For (iii) use ratio test
    – Naweed G. Seldon
    Nov 25 at 14:31






  • 2




    Please limit yourself to one question per post; three questions in one post makes it too broad.
    – amWhy
    Nov 25 at 14:35
















-1














Good day dear community,



I have to examine the following series for covergence and proof it.



(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$



(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$



My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$



My approach to (ii)



$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$



$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$



$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.



My approach to (iii)



And here I have no idea, whatsoever.






limits proof-verification convergence







share|cite|improve this question






















  • For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
    – Nodt Greenish
    Nov 25 at 14:27












  • For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
    – xbh
    Nov 25 at 14:27








  • 1




    $(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
    – Galc127
    Nov 25 at 14:29










  • For (iii) use ratio test
    – Naweed G. Seldon
    Nov 25 at 14:31






  • 2




    Please limit yourself to one question per post; three questions in one post makes it too broad.
    – amWhy
    Nov 25 at 14:35














-1












-1








-1







Good day dear community,



I have to examine the following series for covergence and proof it.



(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$



(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$



My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$



My approach to (ii)



$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$



$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$



$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.



My approach to (iii)



And here I have no idea, whatsoever.






limits proof-verification convergence







share|cite|improve this question













Good day dear community,



I have to examine the following series for covergence and proof it.



(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$



(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$



My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$



My approach to (ii)



$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$



$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$



$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.



My approach to (iii)



And here I have no idea, whatsoever.






limits proof-verification convergence




limits proof-verification convergence






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asked Nov 25 at 14:24









SacredScout

487




487












  • For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
    – Nodt Greenish
    Nov 25 at 14:27












  • For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
    – xbh
    Nov 25 at 14:27








  • 1




    $(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
    – Galc127
    Nov 25 at 14:29










  • For (iii) use ratio test
    – Naweed G. Seldon
    Nov 25 at 14:31






  • 2




    Please limit yourself to one question per post; three questions in one post makes it too broad.
    – amWhy
    Nov 25 at 14:35


















  • For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
    – Nodt Greenish
    Nov 25 at 14:27












  • For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
    – xbh
    Nov 25 at 14:27








  • 1




    $(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
    – Galc127
    Nov 25 at 14:29










  • For (iii) use ratio test
    – Naweed G. Seldon
    Nov 25 at 14:31






  • 2




    Please limit yourself to one question per post; three questions in one post makes it too broad.
    – amWhy
    Nov 25 at 14:35
















For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27






For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27














For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27






For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27






1




1




$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29




$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29












For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31




For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31




2




2




Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35




Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35










3 Answers
3






active

oldest

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1














For part $(i)$:



Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$



For part $(iii)$:



Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.



$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$






share|cite|improve this answer



















  • 1




    Siong, even $gesumfrac13$ !
    – Yves Daoust
    Nov 25 at 15:57












  • For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
    – SacredScout
    Nov 27 at 16:22






  • 1




    $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
    – Siong Thye Goh
    Nov 27 at 16:36



















1














ii) grouping even-odd terms,



$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.






share|cite|improve this answer





















  • Isn't there another way to conclude that (ii) converges?
    – SacredScout
    Nov 27 at 16:10



















1














The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed



$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    For part $(i)$:



    Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$



    For part $(iii)$:



    Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.



    $$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$






    share|cite|improve this answer



















    • 1




      Siong, even $gesumfrac13$ !
      – Yves Daoust
      Nov 25 at 15:57












    • For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
      – SacredScout
      Nov 27 at 16:22






    • 1




      $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
      – Siong Thye Goh
      Nov 27 at 16:36
















    1














    For part $(i)$:



    Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$



    For part $(iii)$:



    Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.



    $$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$






    share|cite|improve this answer



















    • 1




      Siong, even $gesumfrac13$ !
      – Yves Daoust
      Nov 25 at 15:57












    • For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
      – SacredScout
      Nov 27 at 16:22






    • 1




      $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
      – Siong Thye Goh
      Nov 27 at 16:36














    1












    1








    1






    For part $(i)$:



    Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$



    For part $(iii)$:



    Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.



    $$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$






    share|cite|improve this answer














    For part $(i)$:



    Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$



    For part $(iii)$:



    Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.



    $$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 25 at 16:00

























    answered Nov 25 at 15:08









    Siong Thye Goh

    98.4k1463116




    98.4k1463116








    • 1




      Siong, even $gesumfrac13$ !
      – Yves Daoust
      Nov 25 at 15:57












    • For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
      – SacredScout
      Nov 27 at 16:22






    • 1




      $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
      – Siong Thye Goh
      Nov 27 at 16:36














    • 1




      Siong, even $gesumfrac13$ !
      – Yves Daoust
      Nov 25 at 15:57












    • For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
      – SacredScout
      Nov 27 at 16:22






    • 1




      $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
      – Siong Thye Goh
      Nov 27 at 16:36








    1




    1




    Siong, even $gesumfrac13$ !
    – Yves Daoust
    Nov 25 at 15:57






    Siong, even $gesumfrac13$ !
    – Yves Daoust
    Nov 25 at 15:57














    For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
    – SacredScout
    Nov 27 at 16:22




    For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
    – SacredScout
    Nov 27 at 16:22




    1




    1




    $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
    – Siong Thye Goh
    Nov 27 at 16:36




    $frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
    – Siong Thye Goh
    Nov 27 at 16:36











    1














    ii) grouping even-odd terms,



    $$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.






    share|cite|improve this answer





















    • Isn't there another way to conclude that (ii) converges?
      – SacredScout
      Nov 27 at 16:10
















    1














    ii) grouping even-odd terms,



    $$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.






    share|cite|improve this answer





















    • Isn't there another way to conclude that (ii) converges?
      – SacredScout
      Nov 27 at 16:10














    1












    1








    1






    ii) grouping even-odd terms,



    $$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.






    share|cite|improve this answer












    ii) grouping even-odd terms,



    $$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 at 16:00









    Yves Daoust

    124k671221




    124k671221












    • Isn't there another way to conclude that (ii) converges?
      – SacredScout
      Nov 27 at 16:10


















    • Isn't there another way to conclude that (ii) converges?
      – SacredScout
      Nov 27 at 16:10
















    Isn't there another way to conclude that (ii) converges?
    – SacredScout
    Nov 27 at 16:10




    Isn't there another way to conclude that (ii) converges?
    – SacredScout
    Nov 27 at 16:10











    1














    The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed



    $$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$






    share|cite|improve this answer


























      1














      The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed



      $$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$






      share|cite|improve this answer
























        1












        1








        1






        The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed



        $$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$






        share|cite|improve this answer












        The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed



        $$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 at 18:35









        gimusi

        1




        1






























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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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            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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            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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