Is $\sum_{n=1}^{\infty} \frac{x}{n^{1+a}}$ converge pointwisely on $R$?

Question

$(1).$For any $a>0$, is $\sum_{n=1}^{\infty} \frac{x}{n^{1+a}}$ pointwisely convergent on $R$? Do I need to consider certain Taylor series?

$(2).$And to prove this series does not converge uniformly on $R$:

Since $||\frac{x}{n^{1+a}}||_{C(R)}=\sup\limits_{R}\frac{|x|}{n^{1+a}} \not\rightarrow 0$ for all $n$. To have unifrom convergence , the sequence ofpartial sums has to converge to zero, so the series does not converge uniformly. Is this proof valid?

Answer 1

For the first question, $\sum_{n=1}^\infty\frac{x}{n^{1+a}}$ converges pointwise to $x\zeta(1+a)$ for $a>0$ for all $x\in \mathbb{R}$. We can show this directly by using, say the integral test.

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For the second, you are correct that if the series converges uniformly, then the general terms of the series must converge to $0$ uniformly. Here, we have

$$\lim_{n\to \infty}\frac{x}{n^{1+a}}=0$$

But, for $\epsilon=1$ and for any $n$ we can find $x=n^{1+a}$ such that

$$\frac{x}{n^{1+a}}=\frac{n^{1+a}}{n^{1+a}}=1=\epsilon$$

which negates the uniform convergence of the general terms to $0$.

Inasmuch as the general terms do not converge uniformly to $0$, the series cannot converge uniformly.