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Prove that this series diverges

VincentP

New member
Feb 17, 2012
7
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Feb 13, 2012
1,704
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

$\chi$ $\sigma$