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$$\text{Prove that }\sum_{n=1}^{\infty}\frac{\ln\left(1+\frac{x}{n}\right)}{n} \text{ converges uniformly for }x\in[0,a],\ a > 0$$

I am having a hard time with this problem. I have seen various proofs that the sum does not converge uniformly on $\mathbb{R}$ and I understand them, but how would I prove uniform convergence on $[0,a]$? What we have mostly used in class is the Weierstrass, Abel and Dirichlet criteria to prove uniform convergence of function series (which, of course, is different from proving uniform convergence for a sequence of functions).

My attempt and logic mostly centered on the Weierstrass criterion: $$\sum_{n=1}^{\infty}\left|\frac{\ln\left(1+\frac{x}{n}\right)}{n}\right| \leq \sum_{n=1}^{\infty}\frac{x}{n^2}, \text{ and we know this sum converges.}$$

However, as far as I understand, this isn't correct, because the convergence is based on the value of $x$, so this is punctual (not uniform) convergence. Any help or hint towards the solution is much appreciated!

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  • $\begingroup$ @Mittens I don't understand how you determined that limit. Isn't it something along the lines of $\frac{0}{\infty}$ because the logarithm tends to $\ln(1)$ and the denominator to $\infty$? $\endgroup$
    – J__n
    Dec 12, 2023 at 19:38
  • $\begingroup$ Never mind, I made a mistake $\endgroup$
    – Mittens
    Dec 12, 2023 at 19:40
  • $\begingroup$ You are almost there, $\sum_{n=1}^{\infty}\frac{x}{n^2} \le a\sum_{n=1}^{\infty}\frac{1}{n^2}$ for $x \in [0, a]$. $\endgroup$
    – Martin R
    Dec 12, 2023 at 19:42
  • $\begingroup$ $1+x/n\leq e^{x/n}$ and so $\frac1n\log(1+x/n)\leq \frac{x}{n^2}\leq a/n^2$ foire all $x\in[0,a]$. Uniform convergence follows immediately. $\endgroup$
    – Mittens
    Dec 12, 2023 at 19:43
  • $\begingroup$ @MartinR of course! Can't believe I did not notice that. Thank you and Mittens for the swift explanations! $\endgroup$
    – J__n
    Dec 12, 2023 at 19:46

1 Answer 1

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Claim 1: For $x\geq0$, $\ln(1+x)\leq x$.

Proof of Claim 1: Let $f:[0,\infty)\rightarrow\mathbb{R}$ be defined by $f(x)=x-\ln(1+x)$. For $x>0$, we have $f'(x)=1-\frac{1}{1+x}>0$. Therefore, $f$ is strictly increasing on $[0,\infty)$. In particular, $f(x)\geq f(0)=0$ and the result follows.


For your problem: For $x\in[0,a]$, we have that $\left|\frac{\ln(1+\frac{x}{n})}{n}\right|\leq\frac{\ln(1+\frac{a}{n})}{n}\leq\frac{a}{n^{2}}$ by Claim 1. Since $\sum_{n=1}^{\infty}\frac{a}{n^{2}}<\infty$, the infinite series $\sum_{n=1}^{\infty}\frac{\ln(1+\frac{x}{n})}{n}$ converges uniformly for $x\in[0,a]$ by Weierstrass $M$-test. (Also, the infinite series converges absolutely).

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