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We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?

I tried using the ratio test, but the limit $$\lim\limits_{n\to\infty}\left((n+1)^3\sin\frac{\pi}{3^{n+1}}\cdot \frac{1}{n^3\sin\frac{\pi}{3^n}}\right)$$ isn't that easy to evaluate.

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    $\begingroup$ use $\sin x\leq x$ $\endgroup$ – dezdichado Apr 6 '16 at 9:02
  • $\begingroup$ The limit is $1 \over 3$ $\endgroup$ – crbah Apr 6 '16 at 9:15
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    $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. :-) Ofc, you probably already know this by now... $\endgroup$ – Simply Beautiful Art Sep 27 '17 at 16:10
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Since $\sin x \le x$ for $x\ge 0$ and $0\le \frac{\pi}{3^n}\le \pi$, $$ 0\le n^3 \sin\frac{\pi}{3^n} \le \frac{n^3 \pi}{3^n} $$ and $\sum_{n=1}^{\infty}\frac{n^3 \pi}{3^n}$ converges by ratio test. By comparison test, given series converges.

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You can evaluate the limit in this way

$$\begin{align} \lim_{n\to\infty} \left( \frac{(n+1)^3}{n^3} \cdot \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}}\right) &= \lim_{n \to \infty} \frac{(n+1)^3}{n^3} \lim_{n \to \infty} \frac{\sin\frac{\pi}{3^{n+1}}}{\sin\frac{\pi}{3^{n}}} \\ &= 1 \cdot \lim_{n \to \infty} \frac{\frac{\pi}{3^{n+1}}}{\frac{\pi}{3^{n}}} \\ &= 1 \cdot \frac{1}{3} \\ &= \frac{1}{3} \end{align}$$

and since the result is less than $1$ the series will converge.

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This is not a full answer

The other answers already dealt with the first part of the question.

However, since the OP also asks about the sum of the series, I have made some attempt to find a closed form. I haven't been able to so far, but there's a way to transform the series to obtain faster convergence.

First let's use the Taylor series for $\sin$:

$$\sum_{n=1}^{\infty} n^3\sin\frac{\pi}{3^n}=\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} n^3 \frac{\pi^{2k+1}}{3^{n(2k+1)}}$$

Now we can exchange the order of summation, since we have a simple geometric like sum for $n$. I'll leave the derivation out (it can be done by repeated differentiation of the general term of the geometric series w.r.t. $x$ and some simple algebra):

$$\sum_{n=1}^{\infty} n^3 x^n=\frac{x(x^2+4x+1)}{(1-x)^4}$$

Thus we obtain:

$$\sum_{n=1}^{\infty} n^3 \frac{1}{3^{(2k+1)n}}=\frac{1 + 4/3^{2k+1} + 1/3^{2(2k+1)}}{3^{2k+1}(1 - 1/3^{2k+1})^4}=3^{2k+1} \frac{3^{2(2k+1)} + 4 \cdot 3^{2k+1} + 1}{(3^{2k+1} - 1)^4}$$

The resulting series do not seem to have any closed form, in fact they can't even be expressed as a finite sum of hypergeometric functions (because of the denominator).

However, for $k>>1$ we can approximately write $3^{2k+1} - 1 \asymp 3^{2k+1}$, then we have:

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} 3^{2k+1} \frac{3^{2(2k+1)} + 4 \cdot 3^{2k+1} + 1}{(3^{2k+1} - 1)^4} \pi^{2k+1}> \\ > \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} \left(\frac{1}{3^{2k+1}}+\frac{4}{9^{2k+1}}+\frac{1}{27^{2k+1}} \right)\pi^{2k+1}=\sin \frac{\pi}{3}+4 \sin \frac{\pi}{9}+\sin \frac{\pi}{27}$$

Now to obtain an approximation to the original series we simply truncate them at some $N>>1$ and subtract the general term of the simple series above. In other words:

$$\sum_{n=1}^{\infty} n^3\sin\frac{\pi}{3^n} \approx \sin \frac{\pi}{3}+4 \sin \frac{\pi}{9}+\sin \frac{\pi}{27}+S_N$$

Where:

$$S_N=\sum_{k=0}^{N} \frac{(-1)^k}{(2k+1)!} \left( \frac{27^{2k+1} + 4 \cdot 9^{2k+1} + 3^{2k+1}}{(3^{2k+1} - 1)^4}-\frac{1}{3^{2k+1}}-\frac{4}{9^{2k+1}}-\frac{1}{27^{2k+1}} \right) \pi^{2k+1}$$


Let's compare the convergence rate.

Numerically we have:

$$\sin \frac{\pi}{3}+4 \sin \frac{\pi}{9}+\sin \frac{\pi}{27}=2.350198891212\dots$$

$$S_2=10.363579165119\dots$$ $$S_3=10.363578662463\dots$$ $$S_4=10.363578663311\dots$$ $$S_5=10.363578663310\dots$$

$$\sin \frac{\pi}{3}+4 \sin \frac{\pi}{9}+\sin \frac{\pi}{27}+S_4=\color{blue}{12.71377755452}3 \dots$$

$$\sin \frac{\pi}{3}+4 \sin \frac{\pi}{9}+\sin \frac{\pi}{27}+S_5=\color{blue}{12.713777554522} \dots$$

Where the correct digits are highlighted, because the actual value of the series is:

$$\sum_{n=1}^{\infty} n^3\sin\frac{\pi}{3^n}=12.713777554522\dots$$

On the other hand, using the original form of the series, we obtain for 20 and 29 terms:

$$\sum_{n=1}^{20} n^3\sin\frac{\pi}{3^n}=\color{blue}{12.71377}3054816\dots$$

$$\sum_{n=1}^{29} n^3\sin\frac{\pi}{3^n}=\color{blue}{12.71377755}3872\dots$$

In other words, we need a lot more terms to obtain the same number of correct digits.


If I have any more results I will update this post.

All the numerical computation was carried out with Wolfram Alpha.

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