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I need to prove that for $a>0$ and $\operatorname{Re}z>1$, $\sum_{n=1}^{\infty}\frac{1}{(a+n)^z}$ converges. Here's where I am.
The ratio test gives $\lim |\frac{(a+n)^z}{(a+n+1)^z}|=\lim |\frac{a+n}{a+n+1}|^z=1$, which tells me nothing.

Hints only, please.

stank mcgank
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  • Your limit in question does not hold, since the limit as $n \to \infty$ is actually $1$. In general, for a rational function $P(n)/Q(n)$, when $\deg(P) = \deg(Q)$, the limit as $n\to \infty$ is the ratio of the leading coefficients of $P,Q$. – PrincessEev Mar 17 '23 at 01:44
  • What makes you think this limit is zero? Anyway, do you know what is $|(a+n)^z|$? – Mark Mar 17 '23 at 01:46
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    Why not try the Cauchy condensation test initially? This approach is quite standard on Riemann zeta and can be used on Hurwitz zeta with some manipulation. – KStarGamer Mar 17 '23 at 01:48
  • Yes, the limit is actually 1. Edited. – stank mcgank Mar 17 '23 at 02:00
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    In general, $|k^z|\neq |k|^z$. The left-hand side is real, the right-hand side need not be. – David P Mar 17 '23 at 02:07
  • I figure I'm not allowed to use the condensation test because it isn't in the textbook. – stank mcgank Mar 17 '23 at 05:03

1 Answers1

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Hint: $$ \left| \frac{1}{(a+n)^z}\right| = \frac{1}{|(a+n)^z|} = \frac{1}{(a+n)^{\operatorname{Re}z}} \le \frac{1}{n^{\operatorname{Re}z}} \, . $$

Martin R
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