Convergence of Series of Functions $\sum_{n = 1}^{\infty}(n + 1)e^{1 - nx}$

Question

I'm learning about series of functions and need some help with this problem:

Given the series of function $\sum_{n = 1}^{\infty}(n + 1)e^{1 - nx}$ show that

(i) it converges pointwise but not uniformly on the interval $(0, +\infty)$;

(ii) it converges uniformly on the interval $(1, +\infty)$.

My work and thoughts:

Since I'm having difficulties showing (i) I'll be explaining my work for (ii).

(ii) We note that $\sum_{n = 1}^{\infty}(n + 1)e^{1 - nx} = e \sum_{n = 1}^{\infty}\frac{(n + 1)}{e^{nx}}$ and let $f_n(x) = \frac{(n + 1)}{e^{nx}}$.

Therefore $f'_n(x) = \frac{-(n + 1)ne^{nx}}{e^{2nx}} = \frac{-(n + 1)n}{e^{nx}} < 0 \ \forall{x} \in (1, +\infty)$.

So $f_n$ is decreasing on the interval $(1, +\infty)$. In other words $f_n$ is bounded from above and we can write $$\forall{x} \in (1, +\infty) : |f_n(x)| \leq f_n(1) = \frac{n + 1}{e^{n}}.$$

It is easy to prove that the series $\sum_{n = 1}^{\infty} \frac{n + 1}{e^{n}} < +\infty$ (the series converges by the Limit Comparaison Test).

Hence, by the Weierstrass M-test, we conclude that the given series $\sum_{n = 1}^{\infty}(n + 1)e^{1 - nx}$ is uniformly convergent on the interval $(1, +\infty)$.

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Is my work correct for (ii)? How do I show that (i) the series of functions converges pointwise but not uniformly on the interval $(0, +\infty)$?

Answer 1

Yes, what you did for (ii) is correct.

For (i), the pointwise convergence follows from the ratio test. If the series would be uniformly convergent on $(0,+\infty)$, then the sequence $\left(f_n\right)_{n\geqslant 1}$ would converge uniformly on this interval. This would imply that $\lim_{n\to +\infty}f_n(1/n)=0$.

Answer 2

Note that $$ \sum (n+1)q^n=\frac1{(1-q)^2} $$ and convergence is uniform for $|q|\le r$ for any $r<1$.

Answer 3

Let $$\:F_n(x)=\sum_{j=1}^{n}f_j(x)\:, \:F(x)=\sum_{j\in\mathbf N}f_j(x),\quad\:f_j(x)=(j+1)e^{1-jx}\:\:\:\&\:\:\:(j,n)\in\mathbf N^2\:\:\:\forall j\le n.$$

We suspect that $$\exists\hat{\large\epsilon}>0\:\:\forall N\in\mathbf N,\:\:\exists(k,n)\in\mathbf N^2\:\:\text{ such that for }\:n\ge k\ge N\implies\sup_{0<x\le1}\left|\sum_{j=k}^nf_j(x)\right|=\:...\\...=\left\|F_n(x)-F_k(x)\right\|_{0<x\le1,\infty}>\large \hat\epsilon.$$

For any $\text{fixed }\:N\in\mathbf N\:$ if we choose $\:k=2^N,n=2^{N+1}-1\:$ then we have : $$\left\|F_n(x)-F_k(x)\right\|_{0<x\le1,\infty}=\sum_{j=2^N}^{2^{N+1}-1}e(j+1)>e\:{(2^{N+1}-1)2^{N+1}-2^N(2^N+1)\over 2}>6=\large \hat\epsilon.$$

By the Cauchy criterion for series of functions, $\:F(x)\:$ fails to converge uniformly on $\:(0,1],\:$ hence the uniform convergence is not applicable on $\:(0,\infty).$