Prove that prove that $\sum_{n=1}^{\infty} \frac{\log(1 + \frac{x}{n})}{n}$ is not uniformly convergent on $\mathbb{R}$ by contradiction

Question

I want to prove that $f(x) = \sum_{n=1}^{\infty} \frac{\log(1 + \frac{x}{n})}{n}$ is not uniformly convergent on the whole of $\mathbb{R}$.

My idea was to make the argument that if $f(x)$ were uniformly convergent on $\mathbb{R}$, then $\exists N\in\mathbb{N} $ such that $|f(x) - f_n(x)|<\epsilon $ for any $\epsilon > 0$ and any $x \in \mathbb{R}$ when $N\leq n$. Let $f(x)-f_n(x) = \sum_{k = n+1}^{\infty}\frac{\log(1 + \frac{x}{k})}{k}$. Choose $\epsilon = 1$ and $x = k\cdot e^k$. Then, $\left|\sum_{k = n+1}^{\infty}\frac{\log(1 + \frac{x}{k})}{k}\right|= \left|\sum_{k = n+1}^{\infty}\frac{\log\left(1 + \frac{k\cdot e^k}{k}\right)}{k}\right| > 1$, which contradicts the proposition that we can make $\bigl|f(x) - f_n(x)\bigr|$ smaller than any epsilon by choosing $N$ large enough.

Somehow this argument feels faulty. Can I really choose $x$ in such an arbitrary way? $k$ is after all not a constant, but $x$ might need to be?

Any insights into this would be appreciated.

Answer 1

You have the right idea. Here is another way to use the same idea.

Let $f_n(x)=\sum_{j=1}^n\frac {\log (1+x/j)}{j}$ for $x>-1$ (...so $\log (1+x/j)$ is defined when $j=1$).

Let $f(x)=\lim_{n\to \infty}f_n(x).$

Let $\|f-f_n\|=\sup_{x>-1}|f(x)-f_n(x)|.$

Then directly from the definition, $f_n$ converges uniformly to $f$ on $(-1,\infty)$ iff $\lim_{n\to \infty}\|f-f_n\|=0.$

For $m>1$ let $x_m=m(e^m-1).$ We have $f_m(x_m) -f_{m-1}(x_m)=1.$

For any $n$ we have $$\|f_{n+1}-f\|+\|f-f_n\|\geq$$ $$\geq |f_{n+1}(x_{n+1})-f(x_{n+1})|+|f(x_{n+1})-f_n(x_{n+1})|\geq$$ $$\geq |f_{n+1}(x_{n+1})-f(x_{n+1})+f(x_{n+1})-f_n(x_{n+1})|=$$ $$=|f_{n+1}(x_{n+1})-f_n(x_{n+1})|=1.$$

Therefore $\max (\|f_{n+1}-f\|, \|f-f_n\|)\geq 1/2.$

So for every $n$ there exists $n'\geq n$ (...namely $n'=n$ or $n'=n+1$) such that $\|f-f_{n'}\|\geq 1/2.$ Therefore $\|f-f_n\|$ does not converge to $0$ as $n\to \infty.$

Answer 2

Firstly, the infinite series is not well-defined for all $x\in\mathbb{R}$ and $n$ should not start at $n=0$ (otherwise, zero will appear in the denominator). If we let $x$ take negative value, $\ln(1+\frac{x}{n})$ may be un-defined.

Let me correct the question to: Prove that the infinite series $$ \sum_{n=1}^{\infty}\frac{\ln(1+\frac{x}{n})}{n} $$ converges pointwisely but not uniformly on $[0,\infty)$.

Proof: We firstly prove that the infinite series converges pointwisely for each $x\in[0,\infty)$. Let $x\in\mathbb{R}$ be fixed. Note that $\frac{\ln(1+\frac{x}{n})}{n}\geq0$ and $$ \sum_{n=1}^{\infty}\frac{\ln(1+\frac{x}{n})}{n}\leq\sum_{n=1}^{\infty}\frac{x}{n^{2}}<\infty. $$ Here, we have used the fact $\ln(1+x)\leq x$ for all $x\in[0,\infty)$. By comparison test, the infinite series converges. Denote $S(x)=\sum_{n=1}^{\infty}\frac{\ln(1+\frac{x}{n})}{n}$.

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Next, we show that the infinite series does not coverge uniformly to $S(x)$ on $[0,\infty)$. Denote the partial sum by $S_{n}(x)$. That is, $S_{n}(x)=\sum_{k=1}^{n}\frac{\ln(1+\frac{x}{k})}{k}$. Prove by contradiction. Suppose the contrary that $S_{n}\rightarrow S$ uniformly on $[0,\infty)$. Then for $\epsilon=0.1$, there exists $N$ such that $\sup_{x\in[0,\infty)}|S_{m}(x)-S_{n}(x)|<\epsilon$ whenever $m,n\geq N$. Observe that $\frac{\ln(1+\frac{x}{N+1})}{N+1}\rightarrow\infty$ as $x\rightarrow\infty$, so there exists $x_{0}\in[0,\infty)$ such that $\frac{\ln(1+\frac{x_{0}}{N+1})}{N+1}>1$. Finally, observe that $$ 1<\frac{\ln(1+\frac{x_{0}}{N+1})}{N+1}=\left|S_{N+1}(x_{0})-S_{N}(x_{0})\right|\leq\sup_{x\in[0,\infty)}|S_{N+1}(x)-S_{N}(x)|<\epsilon, $$ which is a contradiction.

Answer 3

Useful general result: If $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on a set $E,$ then $\sup_{E} |f_n| \to 0$ as $n\to \infty.$ In your problem, take $E=[0,\infty),$ $f_n(x) = (\ln (1+x/n))/n.$ Then $\sup_{E} |f_n| =\infty$ for each $n.$ Thus uniform convergence in your problem fails, bigtime.