1
$\begingroup$

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 $0<a\in \mathbb{R}$. 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 $f'(x) = \frac{-e^{-x}}{1-e^{-x}}$.

$\endgroup$
1
  • 3
    $\begingroup$ With your reputation you should be able to show some work. $\endgroup$ Jan 15 at 0:11
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$f'(x) = \frac{-e^{-x}}{1-e^{-x}}$$

$\endgroup$
7
  • $\begingroup$ At $4)$ showing that $f_n'(x)$ converges uniformly to $g:=f'$ is equivalent to show that $f$ is differentiable? $\endgroup$
    – Mary Star
    Jan 15 at 8:37
  • $\begingroup$ 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? $\endgroup$
    – Mary Star
    Jan 15 at 9:26
  • 1
    $\begingroup$ @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 $f'$. $\endgroup$ Jan 15 at 16:44
  • $\begingroup$ 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? $\endgroup$
    – Mary Star
    Jan 15 at 16:56
  • 1
    $\begingroup$ @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 $f_n'(y)\to f'(y)$. $\endgroup$ Jan 16 at 4:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.