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I want to show that

$$\lim\limits_{n\to \infty}\sum_{k=1}^{\infty}2^{-k}\sin(k/n)=0$$

I first thought if I can change the order of limit, it can be easy to show that.

But I found that there exists counter example here.

I know that $\sum_{k=1}^{\infty}2^{-k}\sin(k/n)$ absolutely converges, but how to show the result?

Community
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user347643
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  • For fixed $n$, you can find an explicit formula for the sum by using the fact that $\sin(k/n)$ is the imaginary part of $e^{ik/n}$. – André Nicolas Jun 14 '16 at 04:31
  • Neither complex numbers nor truncating sums are necessary, simply use that for every real $x$, $$|\sin x|\leqslant|x|,$$ hence $$\left|\sum_{k=1}^{\infty}2^{-k}\sin(k/n)\right|\leqslant\sum_{k=1}^{\infty}2^{-k}\left|\sin(k/n)\right|\leqslant{}{}{}{}{}{}{}\sum_{k=1}^{\infty}2^{-k}k/n={}{}{}{}{}\frac1n\sum_{k=1}^{\infty}2^{-k}k=\frac2n\to0$$ – Did Jun 14 '16 at 06:06

4 Answers4

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$$\sum_{k\geq 1}2^{-k}\sin(k/n) = \text{Im}\sum_{k\geq 1}\left(\frac{e^{i/n}}{2}\right)^k = \text{Im}\left(\frac{e^{i/n}}{2-e^{i/n}}\right) $$ and $\frac{e^{i/n}}{2-e^{i/n}}\to 1$ as $n\to +\infty$, hence the wanted limit is clearly zero.

Jack D'Aurizio
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$2^{-k} \cdot \sin(k/n)$ is uniformly bounded by $2^{-k}$. So, it's fine to switch the limit's ordering. In another way,

$$|2^{-k} \cdot \sin(k/n)| \le 2^{-k}$$

So by the Dominated Convergence Theorem, the order of the limit can be interchanged with the summation.

Zach466920
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For any $n\in\mathbb N$, write $$\left|\sum_{k=1}^{\infty}2^{-k}\sin(k/n)\right|\leq\left|\sum_{k=1}^n2^{-k}\sin(k/n)\right|+\left|\sum_{k=n+1}^{\infty}2^{-k}\sin(k/n)\right|\\ \leq\sum_{k=1}^n2^{-k}|\sin(k/n)|+\sum_{k=n+1}^{\infty}2^{-k}|\sin(k/n)|\\ \leq \sum_{k=1}^n\frac{2^{-k}k}{n}+\sum_{k=n+1}^{\infty}2^{-k}\\ \leq\frac{1}{n}\sum_{k=1}^n2^{-k}k+2^{-n}\\ \leq \frac{1}{n}\sum_{k=1}^{\infty}2^{-k}k+2^{-n}.$$ Since the series $\sum_{k=1}^{\infty}2^{-k}k$ converges, letting $n\to\infty$ shows the result.

detnvvp
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We can use the identity $sin(x)<x$ if ${x\to 0}$, So this expression can be written as $$L=\lim\limits_{n\to \infty}\sum_{k=1}^{n}2^{-k}\sin(k/n)<\lim\limits_{n\to \infty}\sum_{k=1}^{n}2^{-k}(k/n)$$ Now it's very easy to compute the summation of $\sum_{k=1}^{n}2^{-k}(k/n)$ . This will be equal to $$S=2-\frac{n+2}{2^n}$$ I have proved this summation before Prove that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $ Ckeck this out. Now, by sandwitch or squeeze theoram, $$L=\lim\limits_{n\to \infty}\sum_{k=1}^{n}2^{-k}(k/n)=\lim_{n\to \infty}(\frac{2}{n}-\frac{n+2}{n2^n})$$ So this limit clearly goes to zero as $n\to \infty$. So $$L=0$$ Hence Proved

Vineet Mangal
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