# Pointwise and uniform convergence of sum

For $$x\in \mathbb{R}$$ we have $$f_n(x)=\sum_{k=1}^n\frac{e^{-kx}}{k}$$

1. Show that $$(f_n)_n$$ converges pointwise on $$(0,+\infty)$$.

2. Let $$0. Show that $$(f_n)_n$$ converges uniformly on $$[a,+\infty)$$.

3. Show that $$\displaystyle{f(x)=\lim_{n\rightarrow +\infty}f_n(x)}$$ is continuous on $$(0, +\infty)$$.

4. Show that $$f$$ is differentiable on $$(0, +\infty)$$ with derivative $$f'(x)=\frac{-e^{-x}}{1-e^{-x}}$$.



At 1 do we have to tae the limit $$n\rightarrow +\infty$$ ? But how can we calculate that series?



EDIT :

At question $$4.$$ do we do the following?

Each of the functions $$\{f_n\}$$ is continuously differentiable. It holds that $$f_n'(x)=\sum_{k=1}^n -e^{-kx}$$
Let $$x\in (0,\infty)$$ fixed.

We have that $$g(x)=\lim_{n\rightarrow +\infty}\sum_{k=1}^n -e^{-kx}=-\sum_{k=1}^\infty e^{-kx} = -e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{-e^{-x}}{1-e^{-x}}$$ So $$\{f_n'\}$$ converges pointwise to the function $$g$$ on $$(0,\infty)$$.

Let $$\alpha > 0$$ fixed.

Since $$-e^{-kx}$$ is an increasing function for $$x\in [0, \alpha)$$, then $$-e^{-kx}\leq -e^{-\alpha k}=-(\frac{1}{e^{\alpha}})^k$$.

Since $$\alpha > 0$$, then $$\frac{1}{e^{\alpha}}< 1$$ and so we have $$\sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k<\infty$$ From Weierstrass M-test we get that the series $$\displaystyle{\sum_{k=1}^\infty -e^{-kx}}$$ converges uniformly on $$[0, \alpha)$$ since $$\sum_{k=1}^\infty -e^{-kx} \leq \sum_{k=1}^\infty -e^{-k\alpha}\leq \sum_{k=1}^\infty -\Big(\frac{1}{e^{\alpha}}\Big)^k$$
Since $$\alpha>0$$ is arbitrary, this holds for each $$\alpha$$, and so also for $$\bigcup_{\alpha> 0}[0, \alpha) = (0,\infty)$$.

So we have that $$\{f_n'\}$$ converges uniformly to $$\displaystyle{g(x)=\frac{-e^{-x}}{1-e^{-x}}}$$ that is continuous on $$(0,+\infty)$$.

Then $$f$$ is also continuously differentiable on $$(0,+\infty)$$ and it holds that $$f'=g$$, so $$f'(x) = \frac{-e^{-x}}{1-e^{-x}}$$.

• With your reputation you should be able to show some work. Jan 15 at 0:11

You don't necessarily need to evaluate the series to prove that it converges. To prove that $$( f_n(x) )_n$$ converges, you could prove that it is an increasing and bounded sequence.

For $$1)$$ fix $$x\in (0,\infty)$$ and notice that $$\{f_n(x)\}$$ is a strictly increasing sequence of real numbers. Since $$0\leq f_n(x)=\sum_{k=1}^n \frac{e^{-kx}}{k}\leq \sum_{k=1}^\infty e^{-kx} = e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{e^{-x}}{1-e^{-x}}$$ We know that this geometric series converges because $$e^{-x}<1$$ since $$x\in (0,\infty)$$. Thus $$\{f_n(x)\}$$ is an increasing sequence of real numbers which is bounded above, thus there exists some number $$f(x)$$ such that $$f_n(x)\to f(x)$$. Hence $$\{f_n\}$$ converge pointwise to some function $$f$$ on $$(0,\infty)$$.

For $$2)$$ let $$\alpha > 0$$ be fixed. Since $$e^{-kx}$$ is a decreasing function for $$x\in [\alpha, 0)$$, then $$e^{-kx}\leq e^{-\alpha k}=(\frac{1}{e^{\alpha}})^k$$. Since $$\alpha > 0$$, then $$\frac{1}{e^{\alpha}}< 1$$ so that the geometric series $$\sum_{k=1}^\infty \Big(\frac{1}{e^{\alpha}}\Big)^k<\infty$$ Hence by the Weierstrass M-test we get that the series $$\sum_{k=1}^\infty \frac{e^{-kx}}{k}$$ converges uniformly on $$[\alpha, 0)$$ via $$\sum_{k=1}^\infty \frac{e^{-kx}}{k} \leq \sum_{k=1}^\infty \frac{e^{-k\alpha}}{k}\leq \sum_{k=1}^\infty \Big(\frac{1}{e^{\alpha}}\Big)^k$$

For $$3)$$, use that each of the functions $$\{f_n\}$$ is continuous. Fix $$\alpha>0$$. By $$2)$$ we know that $$f_n\to f$$ uniformly on $$[\alpha,\infty)$$. Now any sequence of continuous functions will converge uniformly to a continuous function. Thus $$f$$ must be a continuous function on $$[\alpha, \infty)$$. This holds for any $$\alpha > 0$$. Thus $$f$$ is a continuous function on $$\bigcup_{\alpha> 0}[\alpha, \infty) = (0,\infty)$$.

For $$4)$$ observe that $$f_n'(x)=\sum_{k=1}^n -e^{-kx}$$ You can use the same argument we used above to show that $$\{f_n'\}$$ converge uniformly to some function $$g(x)=-\sum_{k=1}^\infty e^{-kx} = -e^{-x}\sum_{k=0}^\infty e^{-kx}=\frac{-e^{-x}}{1-e^{-x}}$$ which is continuous on $$(0,\infty)$$. Since we can find a point $$x_0$$ such that $$f_n(x_0)\to f(x_0)$$, then the function sequence of functions $$\{f_n\}$$ converge uniformly to a differentiable function $$f$$ where $$f'=g$$. Thus $$f'(x) = \frac{-e^{-x}}{1-e^{-x}}$$

• At $4)$ showing that $f_n'(x)$ converges uniformly to $g:=f'$ is equivalent to show that $f$ is differentiable? Jan 15 at 8:37
• I haven't really understood the part "Since we can find a point $x_0$ such that $f_n(x_0)\to f(x_0)$, then the function sequence of functions $\{f_n\}$ converge uniformly to a differentiable function $f$ where $f'=g$." Why do show that $f_n(x_0)\to f(x_0)$ and then we get a result about the derivative? I got stuck right now. Could you explain that further to me? Jan 15 at 9:26
• @MaryStar Its part of a theorem then people cite when they want to show that a sequence of functions converge uniformly to a differentiable function. For $4)$ we know that there is a function $g$ that $f_n'$ converge uniformly to. The theorem that I quote states that the function $g$ is in fact $f'$. Jan 15 at 16:44
• So first we want to show that $\{f_n'\}$ converges uniformly to $g$, right? Do we prove that as in question $2)$ ? But in question $2)$ we had an interval $[a, +\infty)$. What do we do in this case? Jan 15 at 16:56
• @MaryStar We know that it converges uniformly and has all the properties for $[a,\infty)$ for any $a>0$. Take the union of all $[a,\infty)$ which is $(0,\infty)$. You can think of it like this. Fix $y\in (0,\infty)$. We can find an $a>0$ so that $y\in [a,\infty)$. Then $f_n(y)\to f(y)$ and $f_n'(y)\to f'(y)$. Jan 16 at 4:42