# Limit $(t-1)\zeta(t)$ as $t\rightarrow 1^+$

Show that $\lim_{t\rightarrow 1^+}(t-1)\zeta(t)=1$.

For $t>1$, we can use the definition $\zeta(t)=\sum_{n=1}^\infty \dfrac{1}{n^t}$, so it is approximately $\int_1^\infty \dfrac{1}{x^t}dx$.

How can this lead to the limit?

-
Did you try rewriting this as some fraction, obtaining an indeterminate form them applying l'Hopitals rule? – mathematics2x2life Dec 14 '13 at 19:31

Hint

For $n\ge1$ we have $$\int_n^{n+1}\frac{dx}{x^t}\le \frac{1}{n^t}$$ so $$\frac{1}{t-1}=\int_1^{\infty}\frac{dx}{x^t}\le \sum_{n=1}^\infty \frac{1}{n^t}$$

and prove by the same method that $$\sum_{n=1}^\infty \frac{1}{n^t}\le 1+\frac{1}{t-1}$$ and conclude by the squeeze theorem.

-

The following series representation is useful for your purpose

$$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n \; (s-1)^n,$$

where $\gamma_n$ is the Stieltjes constant.

$$\gamma_n = \lim_{m \rightarrow \infty} {\left(\left(\sum_{k = 1}^m \frac{(\ln k)^n}{k}\right) - \frac{(\ln m)^{n+1}}{n+1}\right)}.$$

-
This is surely overkill, though. – Ryan Reich Dec 14 '13 at 19:39