# How to prove that $\sum_{n=1}^\infty \left(e^{\frac 1 n} - e^{\frac 1 {n+2}}\right)$ converges?

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?

• Probably calculating the telescoping partial sums is the most simple way here.
– Aig
Commented Feb 24 at 19:20

$$f(x) = e^x$$ 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 $$\sum 1/n^2$$.

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

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

• @MartinR Yes I noticed that! I was indeed fixing but the site is currently not working well!
– user
Commented Feb 24 at 20:06
• The same estimate follows also from the well-known inequality $1+x \le e^x \le 1/(1-x)$. Commented Feb 24 at 20:11
• @MartinR Nice idea this one also!
– user
Commented Feb 24 at 20:13

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.

• 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. Commented Feb 24 at 21:23
• @CalumGilhooley yes, I abuse the symbol $\sim$ Commented Feb 24 at 21:28

Another way, using that by standard limit $$\frac{e^x-1}x\to 1$$ we have

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