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I've got in my assignment to show if the following series converges or diverges. $$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$

Attempt: \begin{align*} \sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) &=\sum_{n=1}^\infty \left( 1 + \frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...-1-\frac{1}{n}\right)\\ &=\sum_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)\\ &=\sum_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right) \end{align*}

At this point I'm lost. I tried using D'Alambert as follows:

$$\lim \frac{a_{n+1}}{a_n}=\lim\frac{\sqrt[n+1]{e}-1-\frac{1}{n+1}}{\sqrt[n]{e}-1-\frac{1}{n}}$$

which I tried to simplify with basic limit laws (hopefully correctly):

$$\lim\frac{\sqrt[n+1]{e}-1}{\sqrt[n]{e}-1} = 1$$

I don't know where to go from here. Thank you for all your help in advance.

rtybase
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Samuel Novelinka
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    $e^{1/n}-1-\frac{1}{n}$ is positive and bounded by $\frac{C}{n^2}$ for any $n\geq 1$, hence the given series is convergent by comparison / the p-test. – Jack D'Aurizio Jan 10 '18 at 13:58
  • Also $$\sum_{m\geq 2}\frac{\zeta(m)}{m!} = \int_{0}^{+\infty}\frac{-x+\sqrt{x},I_1(2\sqrt{x})}{x(e^x-1)},dx. $$ – Jack D'Aurizio Jan 10 '18 at 14:02
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    Using the mean value theorem, there exists a number $\xi \in (0,1/n)$

    $$e^{1/n}=1+\frac1n +\frac12 ,e^{\xi}\left(\frac{1}{n}\right)^2$$

    Hence, we see that since $n\ge 1$

    $$\left|e^{1/n}-1-\frac1n\right|\le \frac{e}{2n^2}$$

     – Mark Viola Jan 10 '18 at 14:22
  • @JackD'Aurizio,This integral with BesselI has closed form solution ? – Mariusz Iwaniuk Jan 10 '18 at 15:26
  • @MariuszIwaniuk: no, but it is very simple to approach numerically, since the integrand function is pretty close to $\frac{1}{2},\exp\left[-\frac{x}{3}-\frac{x^2}{24}\right].$ The value of the wanted series is just a bit larger than one. – Jack D'Aurizio Jan 10 '18 at 15:40
  • You have committed a pretty nasty typo $\sum\limits_{n=1}^\infty \left(\frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right) =\sum\limits_{n=1}^\infty \left(\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...\right)$. Where is the extra $\frac{1}{n}$ coming from? Remove it and you have your answer almost obviously! – rtybase Jan 10 '18 at 17:04

2 Answers2

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Just use comparison test. Using this Proving Schwarz derivative $\frac{f''(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule? you have $$\lim_{n\to\infty}{n^2}\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) =\lim_{x\to 0}\frac1{x^2}\left(e^x -1 -x\right) =\lim\limits_{x\to 0} \frac{\frac{e^x -1}{x}-1}{x}= \frac12$$

Hence, for n large enough we have $$ \left|n^2\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) -\frac1{2}\right|\le \frac14$$

That is $$ \frac1{4n^2}\le \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) \le \frac3{4n^2}$$ But $$\sum_{n=1}^\infty\frac1{n^2}<\infty$$ That is $$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right)$$ converges too

Matthew Towers
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Guy Fsone
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Thank you for the comment. By using the comparison test I easily proved the sum converges.

$$\sum_{n=1}^\infty (\frac{1}{n} + \frac{1}{2!n^2}+\frac{1}{3!n^3}+...) \le \sum_{n=1}^\infty \frac{e}{n^2}$$

Samuel Novelinka
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