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I have the following series:

$$\sum\limits_{n=1}^\infty\left(e^\frac{1}{n} - e^\frac{1}{n+2}\right)$$

I know that this series converges to $e + \sqrt{e} - 2$ (I found its sum using $S = \lim\limits_{n \to \infty} S_n$, where $S_n$ is a partial sum of $n$ elements).

But is there any way to prove that the series converges without finding the actual sum?

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    $\begingroup$ Probably calculating the telescoping partial sums is the most simple way here. $\endgroup$
    – Aig
    Feb 24 at 19:20

4 Answers 4

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$f(x) = e^x$ satisfies $f'(x) = e^x < e$ for $0 < x < 1$, so that by the mean-value theorem $$ 0 < e^\frac{1}{n} - e^\frac{1}{n+2} < e \left( \frac 1n - \frac{1}{n+2}\right) = \frac{2e}{n(n+2)} < \frac{2e}{n^2} \, . $$ This shows that the series converges by comparison with $\sum 1/n^2$.

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We have that

$$ \left(1+\frac1n\right)^n\le e \le \left(1+\frac1n\right)^{n+1}$$

and then

$$e^\frac{1}{n} - e^\frac{1}{n+2}\le \left[\left(1+\frac1{n-1}\right)^n\right]^\frac1n-\left[\left(1+\frac1{n+2}\right)^{n+2}\right]^\frac1{n+2} =\frac1{n-1}-\frac1{n+2}=\frac3{(n-1)(n+2)}$$

therefore the series converges.


As noticed by MartinR in the comments, the same estimate follows also from the well-known inequality $1+x \le e^x \le 1/(1-x)$, indeed

$$e^\frac{1}{n} - e^\frac{1}{n+2}\le \frac1{1-\frac1n}-1-\frac{1}{n+2}=\frac3{(n-1)(n+2)}$$

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  • $\begingroup$ @MartinR Yes I noticed that! I was indeed fixing but the site is currently not working well! $\endgroup$
    – user
    Feb 24 at 20:06
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    $\begingroup$ The same estimate follows also from the well-known inequality $1+x \le e^x \le 1/(1-x)$. $\endgroup$
    – Martin R
    Feb 24 at 20:11
  • $\begingroup$ @MartinR Nice idea this one also! $\endgroup$
    – user
    Feb 24 at 20:13
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At infinity, using Taylor expansion, you have: $$e^{\frac1n}-e^{\frac1{n+2}}\sim 1+\frac1n-1-\frac1{n+2}=\frac{2}{n(n+2)}\sim\frac2{n^2} $$ Hence the series is convergent.

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    $\begingroup$ Do you mean, using the Taylor (Maclaurin) expansion of $e^x$ at $x=0$? If that is what you mean, then the argument doesn't seem to be valid, because it doesn't follow from $e^{1/n}\sim1+1/n$ and $e^{1/(n+2)}\sim1+1/(n+2)$ that $e^{1/n}-e^{1/(n+2)}\sim1/n-1/(n+2),$ even though the conclusion happens to be true. @user's second answer could perhaps be considered as a corrected version of your argument. Alternatively, you could abandon the use of $\sim,$ and write instead $e^{1/n}=1+1/n+O(1/n^2),$ etc. $\endgroup$ Feb 24 at 21:23
  • $\begingroup$ @CalumGilhooley yes, I abuse the symbol $\sim$ $\endgroup$ Feb 24 at 21:28
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Another way, using that by standard limit $\frac{e^x-1}x\to 1$ we have

$$e^\frac{1}{n} - e^\frac{1}{n+2}=e^\frac{1}{n+2}\;\frac{e^\frac{2}{n(n+2)}-1}{\frac{2}{n(n+2)}}\;\frac{2}{n(n+2)}\sim \frac{2}{n(n+2)}$$

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