Convergence of series of $1/n^x$ - pointwise and uniformly,

Question

Consider the series $$\zeta(x) = \sum_{n\ge 1}\frac {1}{n^x}.$$

For which $x \in[0,\infty)$ does it converge pointwise? On which intervals of $[0,\infty)$ does it converge uniformly?

My work:

I think that I can state (without proof, since this goes back to Calc II?) that we have pointwise convergence for $x>1$.

For uniform convergence, we use the Weierstrass M-Test, and see pretty easily that our series of functions in x is bounded by a series of constants:

$$|\frac{1}{n^x}| \le \frac{1}{n^2}$$

for each n, and for all $x \in [2,\infty)$.

But the series of constants, $\frac{1}{n^2}$, converges, and so our original series in $x$ converges uniformly in the set $[2,\infty)$.

Is this ok? If yes, then my question is concerning the set $(1,2)$, i.e., the numbers greater than $1$ but less than $2$. We still get pointwise convergence in this set - how can I say something about the possibility of uniform convergence in this set? For now, I don't have a convenient bound as I did for the set $[2,\infty)$.

Is there a refined upper bound or another test for uniform convergence that could be useful - or something that I can say to rule out uniform convergence in this smaller set?

Thanks in advance,

Edit: What is the limit as x goes to infinity? That is, what is $$lim_x->\infty,\zeta(x) = \sum_{n\ge 1}\frac {1}{n^x}?$$

By the dominated convergence theorem (for series), noting that $$|\frac{1}{n^x}| \le \frac{1}{n^{1+ \delta}}$$ for all n , and for all x $\in$ [1+$\delta$,$\infty$), and the fact that $\frac{1}{n^{1+ \delta}}$ is summable, we have that $$\lim_{x->\infty}\sum_{n\ge 1}\frac {1}{n^x} = \sum_{n\ge 1} lim_{x->\infty} \frac {1}{n^x}.$$

...what is this limit? Evaluating the limit on the R.H.S., if we fix any n>1, all of these terms are zero. But what about the case where we fix n=1 and evaluate this limit? It seems that we have a $1^{\infty}$ situation.

Answer 1

Hint. Let $\delta>0$, then, for $x \in [1+\delta,+\infty)$, $$ \frac{1}{n^x}\leq \frac{1}{n^{1+\delta}},\quad n=1,2,\ldots $$ Since the series $\displaystyle \sum_{n\ge 1}\frac {1}{n^{1+\delta}}$ is convergent (compare it to the integral $\displaystyle \int_1^{+\infty}\frac {dx}{x^{1+\delta}}$), then $\displaystyle \sum_{n\ge 1}\frac {1}{n^x}$ is uniformly convergent on each $[1+\delta,+\infty)$.

Answer 2

The given series is the zeta Riemann series and it's pointwise convergent on $(1,+\infty)$. Moreover for all $a>1$ we have

$$\frac{1}{n^x}\le \frac1{n^a},\quad \forall x\ge a$$ and since $\sum\frac1{n^a}$ is convergent then we have uniform convergence on every interval $[a,\infty)$. The given series isn't uniformly convergent on $(1,\infty)$. Indeed

$$\sup_{x>1}\sum_{k=n+1}^\infty\frac1{k^x}\ge \sup_{x>1}\sum_{k=n+1}^{2n}\frac1{k^x}\ge\sup_{x>1} \frac n{(2n)^x}=\frac12$$