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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


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.


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:

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


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

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.


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.