# Find the value $\sum_{n=1}^{\infty}(e-(1+\frac{1}{n})^n)$

How to find the following series' value?

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^n\bigg)$$

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how do you know if the series converges? –  Aang Mar 3 '13 at 22:52
It converges because my book makes me find the value. –  Guillermo Mar 3 '13 at 22:56
Dear Ryuichi, If your book told you to jump off a bridge, would you do it? ;-) –  Bruno Joyal Mar 3 '13 at 22:57
Mathematica tells me the sum does not converge. I don't know how realiable this is but at least we know it's not some standard sum –  muzzlator Mar 3 '13 at 23:06

The telescoping series $$e-\left({1+{1\over n}}\right)^n=\sum_{j=1}^n\left({1+{1\over n}}\right)^{j-1}\left[\exp(1/n)-\left({1+{1\over n}}\right)\right] \exp((n-j)/n)$$ shows that $$e-\left({1+{1\over n}}\right)^n\geq n \left[\exp(1/n)-\left({1+{1\over n}}\right)\right]\geq n \,{1\over 2}\left({1\over n}\right)^2 = {1\over 2n}$$ for all $n\geq 1$. Therefore the OP's series diverges by comparison with the harmonic series.

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The sum is diverging as for $n>2$ $$e-\left(1+\frac{1}{n}\right)^n > \frac{1}{n}$$

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Well, for $n>2$ anyway.... –  Byron Schmuland Mar 3 '13 at 23:12
oh thanks, forgot the $n>2$ –  Dominic Michaelis Mar 3 '13 at 23:13

If the book says it converges, how about the value of

$$\sum_{n=1}^{\infty}\bigg(e-\Big(1+\frac{1}{n}\Big)^{n+1/2}\bigg)$$

(which does converge)?

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