Limit of $a_n=(1+ frac{1}{n^2})^n$ as $nrightarrow infty$












0














I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







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  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11
















0














I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







share|cite|improve this question


















  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11














0












0








0


2





I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis







share|cite|improve this question













I want to determine the limit of the sequence: $$a_n=(1+ frac{1}{n^2})^n rightarrow 1$$
I was thinking that I could do this via the squeeze theorem, I definitely know that:
$$
1=(1)^n leq (1+ frac{1}{n^2})^n leq (1 + frac{1}{n})^n$$
Yet as I take the limit, the right boundary is not strict enough, I end up with $e$, is there another clever approach we could take?






calculus real-analysis




calculus real-analysis






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asked Nov 25 at 6:14









Wesley Strik

1,514422




1,514422








  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11














  • 1




    Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
    – Paramanand Singh
    Nov 25 at 16:11








1




1




Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
– Paramanand Singh
Nov 25 at 16:11




Let $x_n=1+(1/n^2)$ then $n(x_n-1)to 0 $ and hence $x_n^nto 1$. See math.stackexchange.com/a/1451245/72031 Alternatively you can use Bernoulli and Squeeze to prove $(1-n^{-4})^{n}to 1,(1-n^{-2})^nto 1$ and divide.
– Paramanand Singh
Nov 25 at 16:11










3 Answers
3






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4














We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






share|cite|improve this answer























  • DeepSea.Very nice.
    – Peter Szilas
    Nov 25 at 6:44










  • Yes indeed, thank you DeepSea, very satisfying.
    – Wesley Strik
    Nov 25 at 7:11



















3














Note that:
$$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






share|cite|improve this answer





























    2














    You could also notice that by expansion,



    $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



    $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



    $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



    And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer























      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11
















      4














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer























      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11














      4












      4








      4






      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.






      share|cite|improve this answer














      We have: $1 < a_n < 3^{1/n}$. I want to add some key facts that here. It is quite well known that $b_n = left(1+frac{1}{n}right)^n$ is monotonically increasing and is bounded above by $3$. Thus $a_n = sqrt[n]{b_{n^2}} < 3^{1/n}$. From this it follows that $a_n to 1$ and is consistent with what you did.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 25 at 6:28

























      answered Nov 25 at 6:18









      DeepSea

      70.8k54487




      70.8k54487












      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11


















      • DeepSea.Very nice.
        – Peter Szilas
        Nov 25 at 6:44










      • Yes indeed, thank you DeepSea, very satisfying.
        – Wesley Strik
        Nov 25 at 7:11
















      DeepSea.Very nice.
      – Peter Szilas
      Nov 25 at 6:44




      DeepSea.Very nice.
      – Peter Szilas
      Nov 25 at 6:44












      Yes indeed, thank you DeepSea, very satisfying.
      – Wesley Strik
      Nov 25 at 7:11




      Yes indeed, thank you DeepSea, very satisfying.
      – Wesley Strik
      Nov 25 at 7:11











      3














      Note that:
      $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






      share|cite|improve this answer


























        3














        Note that:
        $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






        share|cite|improve this answer
























          3












          3








          3






          Note that:
          $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$






          share|cite|improve this answer












          Note that:
          $$left(1+frac1{n^2}right)^n=left(left(1+frac1{n^2}right)^{n^2}right)^{1/n}sim e^{1/n}to_{infty} 1.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 6:40









          farruhota

          18.9k2736




          18.9k2736























              2














              You could also notice that by expansion,



              $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



              $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



              $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



              And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






              share|cite|improve this answer


























                2














                You could also notice that by expansion,



                $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  You could also notice that by expansion,



                  $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                  $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                  $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                  And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.






                  share|cite|improve this answer












                  You could also notice that by expansion,



                  $$lim_{n to infty}bigg(1+frac{1}{n^2}bigg)^n = lim_{n to infty} bigg[ {n choose 0}+{n choose 1}bigg(frac{1}{n^2}bigg)+{n choose 2}bigg(frac{1}{n^2}bigg)^2+…$$



                  $$= lim_{n to infty} bigg[1+frac{n!}{1!(n-1)!n^2}+frac{n!}{2!(n-2)!n^4}+…$$



                  $$= lim_{n to infty}bigg[1+frac{n}{n^2}+frac{n(n-1)}{2 n^4}+…$$



                  And all the terms with $n$ vanish as $n to infty$ since the denominator dominates, leaving $1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 6:37









                  KM101

                  3,958417




                  3,958417






























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