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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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  • $\begingroup$ how do you know if the series converges? $\endgroup$ – Aang Mar 3 '13 at 22:52
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    $\begingroup$ It converges because my book makes me find the value. $\endgroup$ – Guillermo Mar 3 '13 at 22:56
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    $\begingroup$ Dear Ryuichi, If your book told you to jump off a bridge, would you do it? ;-) $\endgroup$ – Bruno Joyal Mar 3 '13 at 22:57
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    $\begingroup$ 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 $\endgroup$ – muzzlator Mar 3 '13 at 23:06
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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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    $\begingroup$ Well, for $n>2$ anyway.... $\endgroup$ – user940 Mar 3 '13 at 23:12
  • $\begingroup$ oh thanks, forgot the $n>2$ $\endgroup$ – Dominic Michaelis Mar 3 '13 at 23:13
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Taylor expansion of $\frac {e^{x} - 1} {x}$ around $x = 0$ is $$1 + \frac {x} {2} + \frac {x^2} {3} + \cdots > 1 + \frac {x} {2}.$$ Put $n = 1/x$ and let $n \to \infty$, we have $$n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ By Minkowski inequality, we have $$e - \left(1 + \frac {1} {n}\right)^n \geqslant n \left(e^{1/n} - \left(1 + \frac {1} {n}\right)\right) > \frac {1} {2n}.$$ But $\sum \frac {1} {2n}$ diverges, then so does your series.

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