# Thread: Convergence of Dirichlet Series

1. ## Convergence of Dirichlet Series

Show that $L(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^s}$ converges for Re(s) > 0.

2. Use the alternating series test.

$\sum_{n=0}^{\infty} (-1)^n a_n$ converges if $\lim_{n \to \infty} a_n \to 0$ and all the $a_n$ are non-negative.

For your example, $a_n = \frac{1}{(2n-1)^s}$.

Can you finish..? (note that your example starts with n=1 not n=0 as in the definition but it won't make a difference).

3. What do you mean by non-negative, I mean the $a_n$ are complex numbers?

4. Originally Posted by EinStone
Show that $L(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^s}$ converges for Re(s) > 0.
Group the terms by pairs : $\sum_{k=1}^\infty \left(\frac{1}{(2(2k-1)+1)^s}-\frac{1}{(2(2k)+1)^s}\right)$, and prove that this series converges absolutely.
(you'll need an asymptotic expansion as $n\to\infty$ to see that : $\frac{1}{x^s}-\frac{1}{(x+2)^s}=\frac{1}{x^s}\left(1-\left(1+\frac{2}{x}\right)^{-s}\right)\sim \frac{2s}{x^{s+1}}$ as $x\to\infty$

, and $|a^s|=a^{{\rm Re}(s)}$ in the end)

5. But the series converges absolutely only for Re(s) > 1 (this is given), but I need to show convergence for Re(s) > 0.

6. Originally Posted by EinStone
But the series converges absolutely only for Re(s) > 1 (this is given), but I need to show convergence for Re(s) > 0.
Read my post more carefully, it does answer the question (notice the $\frac{1}{x^{s+1}}$...). The initial series doesn't converge absolutely, but the one I wrote does (and its convergence is equivalent to that of the initial series, I let you find why).

7. Sorry, but what you are doing is super confusing, I dont get anything -_-.

8. Originally Posted by EinStone
Sorry, but what you are doing is super confusing, I dont get anything -_-.
Fine. Your series is $\frac{1}{1^s}-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\cdots$. I say: this series converges if and only if the series $\left(\frac{1}{1^s}-\frac{1}{3^s}\right)+\left(\frac{1}{5^s}-\frac{1}{7^s}\right)+\cdots$ (the terms are inside parentheses) converges.

Indeed, if $S_n=\sum_{k=1}^n \frac{(-1)^{k+1}}{(2k-1)^s}$ denotes the sequence of partial sums of the initial series, the second series has partial sums $T_n=S_{2n}$. If $S_n\to\ell$ (i.e. if the first series converges) then $T_n$ as well (as a subsequence). And if $T_n\to\ell$, then $S_{2n}\to\ell$ and $S_{2n+1}=S_{2n}+\frac{1}{(2(2n+1)-1)^s}\to \ell$ as well, hence classically $S_n\to\ell$.

Thus, it suffices to prove the convergence of the series where the terms are grouped two by two. And, as a matter of fact, this new series converges absolutely, as we will prove.

A short computation shows $\frac{1}{x^s}-\frac{1}{(x+2)^s}=\frac{1}{x^s}\left(1-\left(1+\frac{2}{x}\right)^{-s}\right)\sim\frac{2s}{x^{s+1}}$ as $x\to\infty$, hence $\left|\frac{1}{x^s}-\frac{1}{(x+2)^s}\right|\sim\frac{2|s|}{x^{{\rm Re}(s)+1}}$. Substituting $x=2(2n+1)-1$ shows that the absolute value of the general term of the new series is asymptotically equivalent to $\frac{2|s|}{(4n)^{{\rm Re}(s)+1}}$. Since ${\rm Re}(s)+1>1$, this series converges (usual Riemann series). Thus the new series converges absolutely, hence it converges, and the inital series as well.

9. Ah now I got the connection, nice thanks!!