# Prove that this series diverges

#### VincentP

##### New member
I need to prove that
$$\sum_{n=0}^{\infty} \left(\exp\left(\frac{n^2+2n}{n^2+1} \right) - e \right)$$
diverges. The solution suggests using the limit comparison test, but since we didn't
cover that in my class I was wondering if there is some other easy way to prove divergence.
Thank you for your help.
Vincent

Last edited by a moderator:

#### MarkFL

Staff member
I would suggest the direct comparison test with the harmonic series:

$$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$$

which is known to diverge. If you can show that:

$$\displaystyle \frac{1}{n}\le\exp\left(\frac{n^2+2n}{n^2+1} \right)-e$$ where $$\displaystyle n\in\mathbb{N}$$

then you will have demonstrated that the series in question diverges. I would try showing that the $n$th term of the harmonic series converges more rapidly to zero than the given series.

#### chisigma

##### Well-known member
I need to prove that
$$\sum_{n=0}^{\infty} \left(\exp\left(\frac{n^2+2n}{n^2+1} \right) - e \right)$$
diverges. The solution suggests using the limit comparison test, but since we didn't
cover that in my class I was wondering if there is some other easy way to prove divergence.
Thank you for your help.
Vincent
Is...

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

... and the series...

$$\sum_{n=0}^{\infty} \frac{2 n -1}{n^{2}+1}$$

... diverges...

Kind regards

$\chi$ $\sigma$