2
$\begingroup$

I have to check for the following series wether in converges: $$\sum^{\infty}_{n=1}\frac{2n+\sum_{k=0}^\infty \frac{1}{k!}(\exp(0)-1)^k}{2n^2-n}$$

I first tried to calculate the first partialsums and it seems that the series diverges, how do I prove that?

Convention : $0^0 = 1$

  • 0
    Are you sure, that the inner sum contains $e^0-1$? This is $0$, therefore the whole inner sum ist just an endless addition of $0$. This would just leave you with a sum over $\frac{2n}{2n^2-n}$, this is similar to the harmonic series – 2017-01-15
  • 0
    @Laray With the usual convention that $0^0=1$ for the $k=0$ term the general term is actually $\frac{2n+1}{2n^2-n}$. – 2017-01-15
  • 0
    Right, I forgot about the convention, I will edit my post. – 2017-01-15
  • 2
    @mathbeing: You are right, i did not look, at the 0-th term. But the argumentation stays the same. – 2017-01-15
  • 0
    @Laray Yes, it was just some nitpicking for the sake of accuracy. – 2017-01-15

1 Answers 1

0

$\require{cancel}$

Hint: The general term of your series is $$ \frac{2n+1}{2n^2-n}=\frac{\cancel{2n-1}}{n\cancel{(2n-1)}}+\frac{2}{2n^2-n}\geq\frac{1}{n}. $$