How to prove that $ \lim_{u \downarrow 1} (u-1) \zeta(u) =1 $?

Question

I would like to prove that $$ \lim_{u \downarrow 1} (u-1) \zeta(u) =1 \quad .$$ However, I am not sure which form of the Riemann-zeta function I ought to pick in order to compute this limit.

I guessed I could use Hasse's definition:

$$ \zeta(u) = \frac{1}{u-1} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{\infty} (-1)^{k} \binom{n}{k}(k+1)^{1-u} \quad , $$ which would lead to $$ \lim_{u \downarrow 1} (u-1) \zeta(u) = \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{\infty} (-1)^{k} \binom{n}{k} $$ (right?). But how do I prove this double series is equal to $1$?

Or should I use another definition/form of Riemann's zeta function to compute this limit?

Answer 1

What happened to the exponent on $(k+1)$? I think it should look like $$\sum_n\frac1{n+1}\sum_k(-1)^k{n\choose k}$$ and the inner sum is zero except for $n=0$. That is because $$\sum_{k=0}^{\infty}(-1)^k{n\choose k}=\sum_{k=0}^n(-1)^k{n\choose k}\\=(1-1)^n$$ by the Binomial Theorem.

Answer 2

Use the representation (for $s>0$)

$$\zeta(s) =s\int_1^\infty \lfloor x\rfloor x^{-1-s}dx = \frac{s}{s-1} + s\int_1^\infty (\lfloor x\rfloor -x)x^{-1-s}dx$$

This representation can be seen as follows:

Let $N\in \mathbb{N}.$ Now

$$s\int_1^{N+1} \lfloor x\rfloor x^{-1-s}dx = s\sum_{n=1}^N {n\int_n^{n+1}x^{-1-s}dx} = s\sum_{n=1}^N n(-\frac{1}{s}((n+1)^{-s}-n^{-s}))$$ $$ = \sum_{n=1}^N nn^{-s} - \sum_{n=1}^N n(n+1)^{-s}$$ $$ = \sum_{n=1}^N n^{-s} - N(N+1)^{-s} \to \zeta(s) \text{ as } N\to\infty. $$

Answer 3

Standard comparison of series (with decreasing terms) and integrals (of decreasing functions) yields, for every real $u\gt1$, $$\frac{1}{u-1}=\int_1^\infty\frac{\mathrm dt}{t^u}\leqslant\zeta(u)\leqslant1+\int_1^\infty\frac{\mathrm dt}{t^u}=\frac{u}{u-1}.$$

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Hint: The function $t\mapsto1/t^u$ is decreasing on $t\geqslant1$ hence:

• For every $n\geqslant1$, $$\frac1{n^u}=\int_n^{n+1}\frac{\mathrm dt}{n^u}\geqslant\int_n^{n+1}\frac{\mathrm dt}{t^u}.$$
• For every $n\geqslant2$, $$\frac1{n^u}=\int_{n-1}^n\frac{\mathrm dt}{n^u}\leqslant\int_{n-1}^n\frac{\mathrm dt}{t^u}.$$