Does the series $\sum_{1}^{\infty } \frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} } $ converges uniformly on $\left [ 0,\infty \right )$?

Question

Does the series $\sum_{1}^{\infty } \frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} } $ converges uniformly on $\left [ 0,\infty \right )$? Prove or disprove it.

My first attempt is try to use Weierstrass M-test:$$\left |\frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} } \right | =\frac{1}{n}e^{-\frac{x}{n} } =\frac{1}{ne^{\frac{x}{n}} }< \frac{1}{ne^{\frac{0}{n}} }=\frac{1}{n}......(1) $$ but unfortunately $\sum_{1}^{\infty } \frac{1}{n} $ is divergent, so (1) tell me nothing.

My second attempt is try to use Dirichlet test:$$\left | \sum_{n=1}^{k} \left ( -1 \right ) ^n \right | < 2 \qquad \forall x\in \left [0,\infty \right ) ...\space this \space condition \space is \space OK$$

Next assume $b(n)=\frac{e^{\frac{-x}{n}}}{n} $, then...$$b'{(n)} =\frac{ \left (e^{-\frac{x}{n}}\times \frac{x}{n^2}\times n \right ) - \left ( e^{-\frac{x}{n}}\times1 \right ) }{n^2}= \frac{e^{-\frac{x}{n}}\times \left ( \frac{x}{n}-1 \right ) }{n^2}$$ sadly, it seems not to decrease monotonically to zero, so this condition has failed.

What should I do in order to test the uniform convergence of this series on $\left [ 0,\infty \right )$?

Answer 1

Your series $\sum_{n=1}^\infty\frac{(-1)^n}ne^{-\frac xn}$ is uniformly convergent on $[0,+\infty),$ as the sum of:

• a convergent series independent of $x:$ $$\sum_{n=1}^\infty\frac{(-1)^n}n=-\ln2$$
• an alternating series which will be proved uniformly convergent: $$\sum_{n=1}^\infty(-1)^{n+1}a_n,\quad\text{with}\quad a_n:=\frac{1-e^{-x/n}}n\ge0.$$

The positive sequence $(a_n)$ converges uniformly to $0$ (since $a_n<\frac1n$) and is moreover decreasing (contrarily to the original sequence $\left(\frac{e^{-\frac xn}}n\right)$), as the product of two positive decreasing sequences: $\left(\frac1n\right)$ and $\left(1-e^{-\frac xn}\right).$

Therefore, the usual criterion applies (uniformly as can be seen from its proof, since the convergence of $(a_n)$ to $0$ is uniform) and the conclusion follows.

Answer 2

The series in question can be written as $$ \sum_{n=1}^{\infty } \frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} } = \sum_{n=1}^{\infty } \left[\frac{e^{-\frac{x}{2n} }}{2n}-\frac{e^{-\frac{x}{2n-1} }}{2n-1}\right]. $$ Therefore, $$ \sum_{n=M}^{\infty } \frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} } =\begin{cases} \sum_{n=M/2}^{\infty } \left[\frac{e^{-\frac{x}{2n} }}{2n}-\frac{e^{-\frac{x}{2n+1} }}{2n+1}\right]&,\quad M/2\in \Bbb N\\ \sum_{n=(M+1)/2}^{\infty } \left[\frac{e^{-\frac{x}{2n-1} }}{2n-1}-\frac{e^{-\frac{x}{2n} }}{2n}\right]&,\quad (M+1)/2\in \Bbb N \end{cases}. $$ We prove uniform convergence for the case of $M/2\in\Bbb N$. The other case is deduced likewise.

For $M/2\in\Bbb N$, using $e^{-\frac{x}{2n}}<e^{-\frac{x}{2n+1}}$ we obtain $$ { \sum_{n=M/2}^{\infty } \left[\frac{e^{-\frac{x}{2n} }}{2n}-\frac{e^{-\frac{x}{2n+1} }}{2n+1}\right] { < \sum_{n=M/2}^{\infty } \left[\frac{e^{-\frac{x}{2n+1} }}{2n}-\frac{e^{-\frac{x}{2n+1} }}{2n+1}\right] \\= \sum_{n=M/2}^{\infty } \frac{e^{-\frac{x}{2n+1} }}{2n(2n+1)} \\\le \sum_{n=M/2}^{\infty } \frac{1}{2n(2n+1)} \\\le \sum_{n=M/2}^{\infty } \frac{1}{(2n-1)(2n+1)} \\=\frac{1}{2M-2} } } $$ and $$ \sum_{n=M/2}^{\infty } \left[\frac{e^{-\frac{x}{2n} }}{2n}-\frac{e^{-\frac{x}{2n+1} }}{2n+1}\right] { = \frac{e^{-\frac{x}{M} }}{M}- \sum_{n=M/2}^{\infty } \left[\frac{e^{-\frac{x}{2n+1} }}{2n+1}-\frac{e^{-\frac{x}{2n+2} }}{2n+2}\right] \\\ge \frac{e^{-\frac{x}{M} }}{M}- \sum_{n=M/2}^{\infty } \frac{e^{-\frac{x}{2n+2} }}{(2n+1)(2n+2)} \\\ge \frac{e^{-\frac{x}{M} }}{M}- \sum_{n=M/2}^{\infty } \frac{1}{(2n+1)(2n+2)} \\\ge \frac{e^{-\frac{x}{M} }}{M}- \sum_{n=M/2}^{\infty } \frac{1}{2n(2n+2)} \\= \frac{e^{-\frac{x}{M} }}{M}- \frac{1}{2M} \ge -\frac{1}{2M}. } $$ Therefore $\left|\sum_{n=M}^{\infty } \frac{\left ( -1 \right )^{n} }{n}e^{-\frac{x}{n} }\right|\le\frac{1}{M}$ for $M/2\in\Bbb N$. A similar approach can be taken for $(M+1)/2\in\Bbb N$, which proves that the series is uniformly convergent.