Uniform convergence for $\sum_{k=1}^{\infty} (-1)^{k}e^{-kx} $

Question

I am experiencing difficulties while investigating uniform convergence for the series $$\sum_{k=1}^{\infty} (-1)^{k}e^{-kx} $$ $(x \in \mathbb{R}) $ and with the aforementioned topic in general.

Here is what I've done so far:

I have found out that the condition $x>0$ is necessary for simple convergence and absolute convergence (for the latter, I have concluded using Cauchy's criterion which involves limit superior).

My approach for uniform convergence was the following: I need to show that the sequence of partial sums is a Cauchy sequence, using the supremum norm. (I think that in order to use this method, the space has to be closed or be a Banach space or something, but I have no idea why this is the case). Now, using $|* | $ as the sup norm for the interval $ x \in ]0, \infty[$, I get:

$$|\sum_{k=1}^{n} (-1)^{k}e^{-kx} - \sum_{k=1}^{m} (-1)^{k}e^{-kx}|$$

$$|\sum_{k=m+1}^{n} (-1)^{k}e^{-kx} | \leq \sum_{k=m+1}^{n} |(-1)^{k}e^{-kx}|$$ Here, I have used the triangle inequality which I think I can use because the series converges on that interval (is this correct?). Now I am stuck, as I cant get rid of the $x$ since 0 is not included.

The mark scheme confuses me even further, as it evaluates the norm of the series (I dont know for what interval, but I think its $[R, \infty[, R > 0$ ) like so:

$$ |\sum_{k=n+1}^{\infty} (-1)^{k}e^{-kx}| = |\frac{(-1)^{n+1}e^{-(n+1)x}}{1+e^{-x}}| = \frac{1}{(1+e^x)e^{nx}}$$

I have no idea how this is done, so an explanation for this part would be really helpful.

The markscheme then concludes that the series converges uniformly on $[R, \infty[, R > 0$.

If someone could clarify the points above (1.the condition for using the Cauchy sequence in the first place (to do with completeness), 2. the triangle inequality part, 3. the actual solution), I would be grateful.

Answer 1

Hint

Let the geometric sum $$S_n(x)=\sum_{k=1}^n(-e^{-x})^k=$$

$$\frac{-e^{-x}(1-e^{-nx})}{1+e^{-x}}$$

for $x>A>0 , |(-1)^ke^{-kx}|<e^{-kA}$ $\implies $ the series is normally and uniformly convergent at $[A,+\infty)$.

for $x>0$,

$$S(x)=\lim_{n\to+\infty} S_n(x)=\frac{-1}{e^x+1}$$

$$\lim_{x\to 0^+}S(x)=\frac{-1}{2}$$

$$\lim_{x\to 0^+} (-1)^ke^{-kx}=(-1)^k$$

but $\sum (-1)^k$ diverges.

By the double limit theorem, the convergence is not uniform at $(0,+\infty)$.

Answer 2

Here is an elementary approach:

Note that for $|z|<1$ we have $\sum_{k=0}^n (-z)^k = {1 - (-z)^{n+1} \over 1+z}$ and so $|{1 \over 1+z} - \sum_{k=0}^n (-z)^k| = {|z|^{n+1} \over |1+z|}$.

Note that for any $n$, we can find $t_n \in (0,1)$ such that ${t^{n+1} \over |1+t|} = {1 \over 4}$.

Also note that if $|z| \le R < 1$, then ${|z|^{n+1} \over |1+z|} \le {R^{n+1} \over 1-R}$, and so convergence is uniform for $|z| \le R$.

Now let $z = e^{-x}$, note that $e^{-x}<1$ for $x >0$ and $\sum_k (-1)^k e^{-kx} = {1 \over 1+ e^{-x}}$.

By choosing $x_n = -\ln t_n$ we see that convergence is not uniform on $(0,\infty)$.

If we restrict $x$ to lie in $[a,\infty)$, with $a>0$, we see that $|e^{-x}| \le e^{-R} <1$, hence convergence is uniform.

Answer 3

Here's another approach: In general, if the series $\sum_k f_k$ converges uniformly on $E,$ then $\sup_E |f_k| \to 0.$ In this problem $|f_k(x)| = e^{-kx}.$ Since $\sup_{[0,\infty)}e^{-kx} = 1$ for all $k,$ we do not have the condition met, so the series does not converge uniformly on $[0,\infty).$

On the interval $[R,\infty),$ where $R>0,$ $\sup_{[R,\infty)}|f_k| = e^{-kR}.$ Since $\sum_k e^{-kR} = \sum_k (e^{-R})^k$ is a convergent geometric series, $\sum (-1)^ke^{-kx}$ converges uniformly on $[R,\infty)$ by the Weierstrass M-test.