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We are considering the series of general term: $(1+\frac{1}{n})^n$

I need to find if this series converges or diverges.

1) The Alembert rule can't be applied since we find the limit equal to 1. 2) I therefore tried rewriting:

$(1+\frac{1}{n})^n= \exp(n\ln(1+\frac{1}{n}))\\= \exp(n(\frac{1}{n}-\frac{1}{2n^2}+o\big(\frac{1}{n^2}\big)\big)\\=\exp(1-\frac{1}{2n}+ O\big(\frac{1}{n^2}\big)\big)$

How can I continue from here ?

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What happens to your last term when you take the limit as $n\to\infty$? – abiessu Oct 12 '13 at 19:22
up vote 2 down vote accepted


$$\left(1+\frac1n\right)^n\xrightarrow[n\to\infty]{}e\neq 0$$

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So in my answer, I just have to take the limit as n does to infinity. And conclude convergence right ? – user43418 Oct 12 '13 at 19:23
Of course not! What is the basic necessary condition for any infinite series to converge, @user43418? – DonAntonio Oct 12 '13 at 19:24
Sorry! DIVERGENT since if a series converges then the sequence's limit is zero. Hence if the limit is nonzero, the series is non covergent ie divergent – user43418 Oct 12 '13 at 19:25
Exactly so, @user43418 – DonAntonio Oct 12 '13 at 19:26

If you really are asking whether $\sum_{n=1}^\infty (1+ \frac{1}{n})^n$ converges, that fact that $(1+ \frac{1}{n})^n > 1$ for all $n \geq 1$ should make that obvious. You don't need to find $\lim_{n \to \infty} (1+ \frac{1}{n})^n$ or even worry about whether it exists.

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The divergence test should tell you the answer.

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