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

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


*Let $0<a\in \mathbb{R}$. Show that $(f_n)_n$ converges uniformly on $[a,+\infty)$.


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


*Show that $f$ is differentiable on $(0, +\infty)$ with derivative $f'(x)=\frac{-e^{-x}}{1-e^{-x}}$.
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At 1 do we have to tae the limit $n\rightarrow +\infty$ ? But how can we calculate that series?
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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 $f'(x) = \frac{-e^{-x}}{1-e^{-x}}$.
 A: 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.
A: 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
$$f'(x) = \frac{-e^{-x}}{1-e^{-x}}$$
