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We have $(1+\frac{c}{n})^n \to e^c$ as $n \to \infty$. Why does a term in $o(\frac{1}{n})$ does not change this asymptotic?

Muzi
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  • Take logarithms? – Angina Seng Jun 08 '19 at 11:24
  • If that is little-$o$ notation, refer to: https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation – zxcvber Jun 08 '19 at 11:28
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    This has been asked before, https://math.stackexchange.com/questions/2285641/on-the-justification-of-the-result-of-a-limit-utilizing-little-o-quick-questio , https://math.stackexchange.com/questions/71826/limit-involving-the-exponential-of-an-expression-plus-a-small-o , https://math.stackexchange.com/questions/855563/convergence-to-exponential-with-order-1-n , https://math.stackexchange.com/questions/2059789/convergence-of-altered-sequence – Calvin Khor Jun 08 '19 at 11:33
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    and so on in https://approach0.xyz/search/?q=%24%5Cleft(1%2B%5Cfrac%7Bc%7D%7Bn%7D%2Bo%5Cleft(%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5Cright)%5E%7Bn%7D%24&p=1 – Calvin Khor Jun 08 '19 at 11:38

2 Answers2

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Suppose we let the $o(1/n)$ term as $g(n)$ ($\in o(1/n)$).

Then by definition, $$\lim_{n\rightarrow\infty} \frac{g(n)}{1/n} = \lim_{n\rightarrow\infty} ng(n) = 0$$

Thus if we look at the limit, $$\lim_{n\rightarrow\infty} (1+\frac{c}{n}+g(n))^n = \lim_{n\rightarrow\infty}\left(1+\frac{c+ng(n)}{n}\right)^{n\cdot \frac{c+ng(n)}{c+ng(n)}} = \lim_{n\rightarrow\infty}\left(1+\frac{c+ng(n)}{n}\right)^{\frac{n}{c+ng(n)}\cdot (c+ng(n))}$$ $$=\exp(\lim_{n\rightarrow\infty}(c+ng(n))) = e^c$$

as desired.

zxcvber
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If $x_n=o(\frac 1 n)$ then $n\log(1+\frac c n+x_n)-n\log(1+\frac c n)=n\log(1+\frac {x_n} {1+\frac c n})$. This behaves like $\frac {nx_n} {1+\frac c n}$ for large $n$. [ Because $\log(1+x)$ behave like $x$ for $ x$ near $0$]. Of course $\frac {nx_n} {1+\frac c n} \to 0$. By squeeze theorem we see that $n\log(1+\frac c n+x_n)-n\log(1+\frac c n) \to 0$. Take exponential to complete the proof.

Kavi Rama Murthy
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