I was trying to solve this problem with a friend but we got stuck.


We claimed that the series does not converge uniformly which implies that it is not uniformly Cauchy, i.e., $\exists\,\epsilon_0>0$ such that $\forall\,k\in \Bbb{N},\,\exists\,n_k,m_k\geq k,\;\exists \,x_k\subset [0,\infty)$ such that \begin{align}\left|s_{n_k}(x_k)- s_{m_k}(x_k)\right|\geq \epsilon_0\end{align}

Now, let $k\in \Bbb{N}$ be given. Take $n_k=k,\,m_k=k+1,\;x_k=\frac{1}{k+1}\subset [0,\infty)$ and $\epsilon_0 \overset{?}{=}.$ Then, \begin{align}\left|s_{n_k}(x_k)- s_{m_k}(x_k)\right|&=\left| \sum^{n_k}_{i=1}x_k e^{-i x_k}-\sum^{m_k}_{i=1}x_k e^{-i x_k}\right|\\&=\left| x_k e^{-(k+1) x_k}\right|\\&=\left| \frac{1}{k+1} e^{-(k+1) \frac{1}{k+1}}\right|\\&=\left| \frac{1}{k+1} e^{-1}\right|\end{align} I'm stuck at this point. Can anyone help out? Thanks!

  • $\begingroup$ @Mark Viola: I don't understand "lack thereof of the sequence". $\endgroup$ – Omojola Micheal Sep 14 '18 at 13:34
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    $\begingroup$ The sequence $x\,\frac{e^{-Nx}}{1-e^{-x}}$ does not uniformly converge to $0$. $\endgroup$ – Mark Viola Sep 14 '18 at 13:39


First write down the truncate of the series: for $x>0, m, p \in \mathbb N^*$, we have $$ \sum_{m+1}^{m+p} x \mathrm e^{-nx} = \mathrm e^{-(m+1)x} x\sum_0^{p-1} \mathrm e^{-nx} = \mathrm e^{-(m+1)x} x \frac {1-\mathrm e^{-px}}{1- \mathrm e^{-x}}. $$ Now clearly we could take $x =1/(m+1)$, then the sum becomes $$ \mathrm e^{-1} \cdot \frac 1{m+1} \cdot \frac {1 - \exp(-p/(m+1))} {1 - \exp(1/(m+1))}. $$ For brevity, we could let $p = m+1$. Now note that $$ \lim_m \frac {\dfrac 1 {m+1}}{1 - \exp\left(-\dfrac 1 {m+1}\right)} = 1 >1/2 $$ so for sufficiently large $m$, the sum is greater than $$ \mathrm e^{-1} (1-\mathrm e^{-1}) \cdot \frac 12. $$

The proof

In other words, first take $m \in \mathbb N^*$ such that $$ \frac {1/(m+1)} {1 - \exp(-1/(m+1))} > \frac 12, $$ then the sum $$ \left.\sum_{m+1}^{2m+2} xe^{-nx}\right|_{x=(m+1)^{-1}} = \mathrm e^{-1} (1-\mathrm e^{-1})\cdot \frac {1/(m+1)} {1 - \exp(1/(m+1))} > \mathrm e^{-1} (1-\mathrm e^{-1}) \cdot \frac 12 =:\varepsilon_0. $$


We know that $xe^{-nx}$ is continuous on $[0,+\infty)$, thus if the series converges uniformly, then the sum must be continuous $[0,+\infty)$. However the sum is [suppose your sum is started from $n =0$] $$ S(x)= \begin{cases} 0, & x = 0,\\ \dfrac x {1- \mathrm e^{-x}}, & x >0, \end{cases} $$ then $S(0^+) = 1 \neq 0$, i.e. $S$ is not continuous on $[0, +\infty )$, hence the series cannot converge uniformly.

  • $\begingroup$ Thanks a lot! Please, let me study the first proof. I understand the second on. I'll then ask questions where necessary and tick! Thanks, once again! $\endgroup$ – Omojola Micheal Sep 14 '18 at 14:31

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