$f_n(x)=\left(\sin{{1} \over {n}}\right) x^n$

pointwise convergence: $|f_n(x)|=\left(\sin{{1} \over {n}}\right) |x|^n \sim {{|x|^n}\over{n}}$ for $n \rightarrow +\infty$

$\sum\limits_{n=1}^{+\infty}{{|x|^n}\over{n}}$ is a power series and it converges in $E=[-1,1)$

uniform convergence: I calculate sup$_E|f_n(x)|=f_n(1)$ so there isn't unif. convergence in E. Can I have convergence in a subset of E?

If I consider interval $E=[a,b], -1<a<b<1$, $\sup_E|f_n(x)|=\max\left\{\sin{{1} \over {n}}|a|^n,\sin{{1} \over {n}}|b|^n\right\}$, general term of convergent series so there is uniform convergence in $[a,b]$?

  • $\begingroup$ it converges in $(-1,1)$ not $[-1,1)$ $\endgroup$ – Thinking Feb 12 at 17:45
  • $\begingroup$ for $x=-1$ is not Leibniz true? $\endgroup$ – Giulia B. Feb 12 at 17:48
  • $\begingroup$ For $x = -1$ it's just the harmonic serie which diverges $\endgroup$ – Thinking Feb 12 at 17:50
  • $\begingroup$ $\sum_{n=1}^{+\infty} (sin{{1} \over {n}}) (-1)^n$ is a Leibniz series so I have pointwise convergence in $x=-1$ $\endgroup$ – Giulia B. Feb 12 at 17:54
  • $\begingroup$ $\sum_{n=1}^\infty (\sin(1/n))(-1)^n \approx -0.551$. $\endgroup$ – irchans Feb 12 at 17:55

The series converges uniformly on any compact interval $[-1,a] \subset [-1,1)$ with $-1$ as the left endpoint by the Dirichlet test.

Note that $\sin \frac{1}{n} \searrow 0$ monotonically and $\sum_{n = 1}^m x^n$ is uniformly bounded for all $m \in \mathbb{N}$ and all $x \in [-1,a]$ with

$$\left|\sum_{n = 1}^m x^n \right| \leqslant \max(1,|a|/(1-|a|)$$

However even though the series converges pointwise for each $x \in [-1,1)$ the convergence is not uniform on any interval $(a,1)$. We can take $a = 0$ with no loss of generality to prove this non-uniform convergence.

Note that $$\sup_{x \in (0,1)} \left|\sum_{k=n+1}^{\infty}x^k \sin \frac{1}{k}\right| > \sup_{x \in (0,1)} \sum_{k=n+1}^{2n}x^k \sin \frac{1}{k} \\> \sup_{x \in (0,1)} nx^{2n}\sin \frac{1}{2n} = \frac{1}{2}\frac{\sin \frac{1}{2n}}{\frac{1}{2n}}$$

Since the RHS converges to $\frac{1}{2}$ as $n \to \infty$ the Cauchy criterion for uniform convergence is violated.

  • $\begingroup$ I think you can change the RHS of the last inequality to $|a|/(1-|a|)$ discarding the max. $\endgroup$ – irchans Feb 12 at 18:10
  • $\begingroup$ So it doesn't converge on the whole $[-1, 1)$. Since I did a big mistake I am wondering how to prove uniform convergence on an open set $(a,b)$ ? If the serie doesn't converge pointwise at the endpoint $a$ and $b$, I don't see a general strategy to prove uniform convergence... Since uniform convergence is a global property, when the interval is open it seems hard to prove uniform convergence $\endgroup$ – Thinking Feb 12 at 18:10
  • $\begingroup$ @Thinking: What exactly is the question now? On a compact interval $[a,b] \subset (-1,1)$ it is easy to see uniform convergence by the Weierstrass test. The Dirichlet test extends it to $[-1,b]$. $\endgroup$ – RRL Feb 12 at 18:17
  • $\begingroup$ @RRL Sorry, but maybe I should make a post to ask this. Here I am not talking about the serie of the OP. If we have a sequence of $f_n : [a,b] \to \mathbb{R}$ such that $\sum f_n$ converges uniformly on $(a,b)$ but doesn't converge pointwise on the endpoints (ie. not on $a$, not on $b$). Then I am wondering what are the strategy to prove that $\sum f_n$ converges uniformly ? We can't use the $M$-test since $\sum f_n$ don't converge on $a$ and $b$... I mean the fact that uniforme convergence is a global property makes it hard since $(a,b)$ is an open set. So how to do when dealing with this ? $\endgroup$ – Thinking Feb 12 at 18:22
  • $\begingroup$ @Thinking: I finished this problem by showing you why it fails to converge on an interval $[a,1)$. To address your general question you would first need to understand what has been shown here. You are also mixing generalities with specific aspects of this problem. " We can't use the M-test since $\sum f_n$ doesn't ...". And apparently the OP is not even interested in what has transpired here! $\endgroup$ – RRL Feb 12 at 18:56

Note that the series $\sum_{n=1}^\infty \sin\left(\frac1n\right)x^n$ fails to converge uniformly on $(-1,1)$. To see this, we can write

$$\sum_{n=1}^\infty \sin\left(\frac1n\right)x^n=\sum_{n=1}^\infty \left(\sin\left(\frac1n\right)-\frac1n\right)x^n+\sum_{n=1}^\infty \frac{x^n}n\tag1$$

The first series on the right-hand side of $(1)$ converges uniformly on $[-1,1]$ since the term $\sin\left(\frac1n\right)-\frac1n=O\left(\frac1{n^3}\right)$. Therefore, it is enough to show that the second series on the right-hand side of $(1)$ fails to converge on $(-1,1)$.

To negate the uniform convergence of $\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x)$ on $(-1,1)$, we choose $\varepsilon=\frac18\log(2)$. Then, we have for any $N\ge1$ and $x=1-\frac1{N+1}\in(0,1)$

$$\begin{align} \left|-\log(1-x)-\sum_{n=1}^N \frac{x^n}{n}\right|=&\left|\sum_{n=1}^\infty \frac{x^n}{n}-\sum_{n=1}^{N} \frac{x^n}{n}\right|\\\\ &=\sum_{n=N+1}^\infty \frac{x^n}{n}\\\\ &\ge \sum_{n=N+1}^{2N+1} \frac{x^n}{n}\\\\ &\ge \left(1-\frac1{N+1}\right)^{2N+1}\sum_{n=N+1}^{2N+1}\frac1n\\\\ &\ge \left(1-\frac1{N+1}\right)^{2N+1}\log(2)\\\\ &\ge \frac18\log(2)\\\\ &=\varepsilon \end{align}$$

And hence the series $\sum_{n=1}^\infty \frac{x^n}{n}$ converges on $[-1,1)$ but fails to converge uniformly on $[-1,1)$.


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.