Evaluate the limit of...











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$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







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  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22















up vote
4
down vote

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$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







share|cite|improve this question




















  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22













up vote
4
down vote

favorite
4









up vote
4
down vote

favorite
4






4





$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits







share|cite|improve this question















$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$




My try:



The limit can be written as follows:



$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$



Evaluate the following series:



$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$



$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$



Then:



$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.



Therefore:



$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$



What to do next?






calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 4 at 15:31









Asaf Karagila♦

301k32422753




301k32422753










asked Dec 4 at 14:18









user605734 MBS

1789




1789








  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22














  • 3




    You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
    – xbh
    Dec 4 at 14:22








3




3




You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
– xbh
Dec 4 at 14:22




You do not need to calculate the series. Also it is not valid to do so first. Try other things. Maybe Cesaro-Stolz theorem.
– xbh
Dec 4 at 14:22










3 Answers
3






active

oldest

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up vote
5
down vote



accepted










HINT



We have that by Stolz-Cesaro



$$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






share|cite|improve this answer




























    up vote
    2
    down vote













    Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
    and
    $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
    Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
    ?






    share|cite|improve this answer






























      up vote
      1
      down vote













      Thanks to user xbh for the hint:



      Using the Stolz-Cesàro theorem, we have:



      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



      $b_n=n^2$



      Two monotone and increasing sequences.



      Apply the theorem to get:



      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



      Then:



      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






      share|cite|improve this answer

















      • 3




        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
        – gimusi
        Dec 4 at 14:55













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      HINT



      We have that by Stolz-Cesaro



      $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



      $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






      share|cite|improve this answer

























        up vote
        5
        down vote



        accepted










        HINT



        We have that by Stolz-Cesaro



        $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



        $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






        share|cite|improve this answer























          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          HINT



          We have that by Stolz-Cesaro



          $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



          $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$






          share|cite|improve this answer












          HINT



          We have that by Stolz-Cesaro



          $$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$



          $$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 at 14:43









          gimusi

          92.3k84495




          92.3k84495






















              up vote
              2
              down vote













              Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
              and
              $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
              Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
              ?






              share|cite|improve this answer



























                up vote
                2
                down vote













                Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                and
                $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                ?






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                  and
                  $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                  Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                  ?






                  share|cite|improve this answer














                  Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
                  and
                  $$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
                  Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
                  ?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 4 at 14:42

























                  answered Dec 4 at 14:36









                  Robert Israel

                  316k23206457




                  316k23206457






















                      up vote
                      1
                      down vote













                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer

















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55

















                      up vote
                      1
                      down vote













                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer

















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55















                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$






                      share|cite|improve this answer












                      Thanks to user xbh for the hint:



                      Using the Stolz-Cesàro theorem, we have:



                      $a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$



                      $b_n=n^2$



                      Two monotone and increasing sequences.



                      Apply the theorem to get:



                      $frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$



                      Then:



                      $lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 4 at 14:48









                      user605734 MBS

                      1789




                      1789








                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55
















                      • 3




                        Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                        – gimusi
                        Dec 4 at 14:55










                      3




                      3




                      Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                      – gimusi
                      Dec 4 at 14:55






                      Note that form here we don't need the extra exponent indeed we have simply $$=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}=frac{n+1}{2n+1}left(1+frac{1}{n+1}right)^{n+1} to frac e 2$$
                      – gimusi
                      Dec 4 at 14:55




















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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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