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

calculusconvergence-divergencereal-analysissequences-and-seriesuniform-convergence

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 $\left [ 0,\infty \right )$?

Best Answer

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.

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