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

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 $[a,b]$?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $[-1,1)$.

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