3
$\begingroup$

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 $(1, +\infty)$.


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

$\endgroup$
0
$\begingroup$

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 $\lim_{n\to +\infty}f_n(1/n)=0$.

$\endgroup$
  • $\begingroup$ Thank you for your help. I'm still having some difficulties understanding the method for (i). Applying the Ratio Test as you proposed I get: $\lim_{n\to +\infty} \frac{(n + 2)}{(n + 1)e^x}$. For a fixed $x > 0$ the limit $L < 1$ so the series converges by the Ratio Test. Is this the correct application? If yes, why does this not imply that the series of functions converges uniformly on the interval $(0, +\infty)$? $\endgroup$ – Von Kar Jan 25 '16 at 23:01
  • $\begingroup$ The ratio test only shows the pointwise convergence. In order to deduce the uniform convergence, we should have something like $\lim_{n\to \infty}\sup_{x\gt 0}f_{n+1}(x)/f_n(x)\lt 1$, which is not the case here. $\endgroup$ – Davide Giraudo Jan 26 '16 at 9:16
  • $\begingroup$ Right! So pointwise convergence of function series is similarly defined by requiring that the sequence of partial sums converges pointwise. Thank you very much for your clear and helpful explanation. $\endgroup$ – Von Kar Jan 26 '16 at 12:47
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\:(0,\infty).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.