I would like to get feedback about next exercise that I've considered, I don't know if it has a good mathematical content, but I would like to know how get a solution.

We consider series of the form $$\sum_{n=1}^\infty\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda\tag{1}$$ where $\lambda$ is a real number and $\zeta(s)$ is the Riemann zeta function.

The exercise that I've created is to find and justify the existence of a $\lambda$ for which our series $(1)$ is convergent.

Question. Can you find a ray of positive real numbers $(\lambda_0,\infty)$ for which our series $$\sum_{n=1}^\infty\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda$$ is convergent when $\lambda_0<\lambda$? I am asking about what work can be done to get the smallest $\lambda_0>0$ as is possible. Many thanks.


Since $$\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)} = \frac{1}{2\sqrt n} + o\left( \frac{1}{\sqrt n}\right)$$

the series converges iff $\lambda \cdot \frac 12>1$ ie $\lambda > 2$.

As noted by Robert Z in his answer, $\zeta(n) = 1 +o(1)$ is not enough to conclude. He showed that $\zeta(n)=1+O(1/2^n)$ (and also that $\zeta(n)=1+1/2^n + o(1/2^n)$).

With this knowledge, $$\begin{aligned} \frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)} &= \frac{\sqrt{n+1}}{1+O(\frac{1}{2^n})}-\frac{\sqrt{n}}{1+O(\frac{1}{2^n})}\\ &= \sqrt n \left(\sqrt{1+\frac 1n}\cdot\left(1+O(\frac{1}{2^n})\right) -1 +O(\frac{1}{2^n}) \right)\\ &=\sqrt n \left( \left[1+\frac{1}{2n}+o\left(\frac 1n\right)\right]\cdot\left(1+O(\frac{1}{2^n})\right) -1 +O(\frac{1}{2^n}) \right)\\ &= \sqrt{n} \left( 1+\frac{1}{2n} +o\left(\frac 1n\right) -1 +O(\frac{1}{2^n}) \right)\\ &= \frac{1}{2\sqrt n} + o\left( \frac{1}{\sqrt n}\right) \end{aligned}$$

  • $\begingroup$ Many thanks I should to study your answer to fill details, but I'm sure that is right. $\endgroup$ – user243301 Apr 13 '18 at 12:10
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    $\begingroup$ @user243301 you only need the crude estimate $\lim_{x\to \infty} \zeta(x) = 1$. $\endgroup$ – Gabriel Romon Apr 13 '18 at 12:17
  • $\begingroup$ Nice Gabriel! (+1) $\endgroup$ – gimusi Apr 13 '18 at 12:23
  • $\begingroup$ OK. Now it is perfect! (+1) $\endgroup$ – Robert Z Apr 13 '18 at 14:09
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    $\begingroup$ @GabrielRomon: Note that $\lim\limits_{n\to\infty}1+\frac{(-1)^n}{\sqrt{n}}=1$, but $\frac{\sqrt{n+1}}{1+\frac{(-1)^{n+1}}{\sqrt{n+1}}}-\frac{\sqrt{n}}{1+\frac{(-1)^n}{\sqrt{n}}}\sim(-1)^n\cdot2$, so the crude estimate is not enough. $\endgroup$ – robjohn Apr 13 '18 at 14:46

In my opinion the fact that $\zeta(n)\to 1$ is not sufficient for our purpose. It is not true that if $(a_n)$ and $(b_n)$ are positive sequences which tend to $1$ then $$\frac{\sqrt{n+1}}{b_{n}}-\frac{\sqrt{n}}{a_n}\sim \frac{1}{2\sqrt n}.$$ Take for example $b_n=\sqrt{1+\frac{1}{n}}$ and $a_n=1$.

Here we need an estimate of how fast $\zeta(n)\to 1$, $$1\leq \zeta(n)=1+\frac{1}{2^n}+\sum_{k=3}^{\infty}\frac{1}{k^n}\leq 1+\frac{1}{2^n}+\int_{2}^{\infty}\frac{dx}{x^n}= 1+\frac{1}{2^n}+\frac{1}{(n-1)2^{n-1}}.$$ Therefore $\zeta(n)=1+O(1/2^n)$ and we may conclude that $$\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\sim \frac{1}{2\sqrt n}.$$

  • $\begingroup$ You're right, my comment to the OP about the estimate needed is wrong. $\endgroup$ – Gabriel Romon Apr 13 '18 at 13:45
  • $\begingroup$ Thanks I've also updated! $\endgroup$ – gimusi Apr 13 '18 at 14:28
  • $\begingroup$ Many thanks Robert. $\endgroup$ – user243301 Apr 13 '18 at 14:29

Note that since, as noted by Robert Z

$$\zeta(n)=1+\frac1{2^n} + o(1/2^n)\implies \frac1{\zeta(n)}\sim1-\frac1{2^n}$$


$$\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\sim \sqrt{n+1}-\sqrt{n} \sim \frac{1}{2\sqrt n}$$


$$\left(\frac{\sqrt{n+1}}{\zeta(n+1)}-\frac{\sqrt{n}}{\zeta(n)}\right)^\lambda \sim \frac12\frac1{n^\frac \lambda 2}$$

which converges for $\frac \lambda 2>1$ and diverges otherwise.

  • $\begingroup$ Many thanks for your details and the details of the other user. Now I see the clear reasoning. $\endgroup$ – user243301 Apr 13 '18 at 12:22
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    $\begingroup$ @user243301 You are welcome, Bye. $\endgroup$ – gimusi Apr 13 '18 at 12:23

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