Show that the sequence $a_n = (1 + frac{1}{n})^n$ is bounded. [duplicate]












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  • Proving $mathrm e <3$

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Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.










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Nov 25 '18 at 10:51


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    Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
    – Rebellos
    Nov 25 '18 at 10:22












  • Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
    – Neymar
    Nov 25 '18 at 10:30
















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This question already has an answer here:




  • Proving $mathrm e <3$

    7 answers




Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.










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Nov 25 '18 at 10:51


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  • 3




    Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
    – Rebellos
    Nov 25 '18 at 10:22












  • Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
    – Neymar
    Nov 25 '18 at 10:30














-2












-2








-2








This question already has an answer here:




  • Proving $mathrm e <3$

    7 answers




Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.










share|cite|improve this question
















This question already has an answer here:




  • Proving $mathrm e <3$

    7 answers




Show that the sequence $$a_n = bigg(1 + frac{1}{n}bigg)^n$$ is bounded.





This question already has an answer here:




  • Proving $mathrm e <3$

    7 answers








real-analysis sequences-and-series convergence upper-lower-bounds






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edited Nov 25 '18 at 10:21









Rebellos

14.4k31245




14.4k31245










asked Nov 25 '18 at 10:15









user619263

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marked as duplicate by José Carlos Santos, Yanko, egreg real-analysis
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Nov 25 '18 at 10:51


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Nov 25 '18 at 10:51


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
    – Rebellos
    Nov 25 '18 at 10:22












  • Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
    – Neymar
    Nov 25 '18 at 10:30














  • 3




    Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
    – Rebellos
    Nov 25 '18 at 10:22












  • Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
    – Neymar
    Nov 25 '18 at 10:30








3




3




Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 '18 at 10:22






Hi, I edited your question format a bit. Welcome to MSE. Please, in future, use Math-Jax to formulate your questions ! Moreover, do you have any thoughts on the given problem ? Do you know the number $e$ ? Maybe you can see a pattern ?
– Rebellos
Nov 25 '18 at 10:22














Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 '18 at 10:30




Prove by induction it is bounded above by 3. For n=1 the result is evident. Next, notice a(n+1)= (n+2)^(n+1)/(n+1)^(n+1)... can you continue?
– Neymar
Nov 25 '18 at 10:30










2 Answers
2






active

oldest

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0














Using the inequality
$$
(1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
$$

and the binomial theorem
we obtain
$$
2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
le
sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
$$






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    0














    Hints:



    $$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$



    $$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$



    Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Using the inequality
      $$
      (1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
      $$

      and the binomial theorem
      we obtain
      $$
      2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
      le
      sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
      $$






      share|cite|improve this answer


























        0














        Using the inequality
        $$
        (1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
        $$

        and the binomial theorem
        we obtain
        $$
        2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
        le
        sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
        $$






        share|cite|improve this answer
























          0












          0








          0






          Using the inequality
          $$
          (1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
          $$

          and the binomial theorem
          we obtain
          $$
          2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
          le
          sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
          $$






          share|cite|improve this answer












          Using the inequality
          $$
          (1+a)^nge 1+na, quad ninmathbb N, ,,age-1,
          $$

          and the binomial theorem
          we obtain
          $$
          2=1+ncdotfrac{1}{n}leleft(1+frac{1}{n}right)^n=sum_{k=0}^nbinom{n}{k}frac{1}{n^k}=sum_{k=0}^nfrac{n!}{k!(n-k)!n^k}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{k!n^k}\
          le
          sum_{k=0}^nfrac{1}{k!}=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+cdots+frac{1}{n!}le 2+frac{1}{2}+frac{1}{2^2}+cdots+frac{1}{2^{n-1}}<3.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 '18 at 10:47









          Yiorgos S. Smyrlis

          62.5k1383162




          62.5k1383162























              0














              Hints:



              $$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$



              $$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$



              Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)






              share|cite|improve this answer


























                0














                Hints:



                $$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$



                $$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$



                Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)






                share|cite|improve this answer
























                  0












                  0








                  0






                  Hints:



                  $$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$



                  $$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$



                  Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)






                  share|cite|improve this answer












                  Hints:



                  $$bigg(1 + frac{1}{n}bigg)^n = {nchoose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+{n choose 3}bigg(frac{1}{n}bigg)^3+... tag{1}$$



                  $$frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+...<frac{1}{2}+frac{1}{4}+frac{1}{8}+...tag{2}$$



                  Try simplifying $(1)$ and letting $n to infty$. See whether it converges to a certain value. You can show whether the series is bounded by using hint $(2)$. (The result should be familiar...)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 10:51









                  KM101

                  4,8861421




                  4,8861421















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