# $\sum\limits_{n=0}^{\infty}\frac{1}{n+\exp(n)}$

I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{1}{n+\exp(n)}$ converges or diverge.

I tried to use the comparison test: $$|\frac{1}{n+\exp(n)}|\leq \frac{1}{\exp(n)} = \frac{1}{\frac{n^n}{n!}} = \frac{n!}{n^n}$$

$\frac{n!}{n^n}$ converge $\Rightarrow$ $\sum\limits_{n=0}^{\infty}\frac{1}{n+\exp(n)}$ converge.

Can somebody please tell me, if this is correct?

• Do you think that $\exp(n) = \frac{n^n}{n!}$? – Najib Idrissi Dec 6 '13 at 19:09
• $\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}$ is the definition of $\exp(x)$. So $\exp(n) = \frac{n^n}{n!}$. Can somebody please tell me, if this is correct? – fear.xD Dec 6 '13 at 19:20
• No, this is not correct. You cannot reuse the index in the sum. $$\exp(n) = \sum_{k=0}^\infty \frac{n^k}{k!}$$ – Najib Idrissi Dec 6 '13 at 19:28

There is a much easier way. Note that $$\dfrac1{n+e^n} < \dfrac1{e^n}$$ Hence, $$\sum_{n=0}^{\infty}\dfrac1{n+e^n} < \sum_{n=0}^{\infty}\dfrac1{e^n} = \dfrac{e}{e-1}$$
Use Quotient test to see that $$n^{3/2}u_n\to 0<\infty, ~~n\to\infty$$
• It is clear that $u_n=O(n^2)$ – Gabriel Romon Dec 6 '13 at 19:19
• I forgot the minus sign $O(n^{-2} )$. I like the bigO notation better, as everybody :) . – Gabriel Romon Dec 6 '13 at 19:35
The transition $1/e^n = \frac{n^n}{n!}$ inaccurately and needs some explanation, what you can do instead is use the fact the $a_n=1/e^n$ is geometric sequence.