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I want to test the convergence of the series. $$ \sum_{n=1}^{\infty} \frac{\sin(n)}{n} \cdot \left(1+\frac{1}{2} + \cdots + \frac{1}{n}\right)$$ My guess is this should diverge, and the below I provide the details

Note that

  • $\displaystyle 1+\frac{1}{2} + \cdots + \frac{1}{n} \geq 1 +\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{n-1}} = 2 \cdot \left(1-\frac{1}{2^{n}}\right)$

  • So from above I conclude that $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n} \cdot \left(1+\frac{1}{2}+\cdots + \frac{1}{n}\right) \geq 2 \cdot \sum_{n=1}^{\infty} \frac{\sin(n)}{n} - \cdot \underbrace{\sum_{n=1}^{\infty} \frac{\sin(n)}{n \cdot 2^{n}}}_{A}$$

  • Now $A$ converges and I think $\displaystyle \sum \frac{\sin(n)}{n}$ diverges hence my given series should diverge. Am I correct in my reasoning?

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    $\begingroup$ No, $\sum \frac{\sin(n)}{n}$ converges by the Dirichlet's test - en.wikipedia.org/wiki/Dirichlet's_test $\endgroup$ – Dennis Gulko Apr 23 '13 at 14:52
  • $\begingroup$ use that $\sum \limits^k sin(n)$ is bounded and sum by parts $\endgroup$ – mike Apr 23 '13 at 14:54
  • $\begingroup$ @DennisGulko Thank you. $\endgroup$ – ra.ss Apr 23 '13 at 14:56
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The series is convergent and, for fun, I'll provide a derivation.

Let's first rewrite your sum as ($H_n$ is the $n$-th harmonic number) : $$\tag{1}S:=\sum_{n=1}^{\infty} H_n\frac{\sin(n)}{n}=\Im\left(\sum_{n=1}^{\infty} H_n\frac{e^{in}}{n}\right)$$

Since $H_n\sim \log(n)\,$ as $\,n\to\infty$ (because the 'harmonic integral' corresponding to the arithmetic sum is the logarithm)

To evaluate $S$ let's use generating functions and suppose that the derivative of $f$ is defined by : $$\tag{2}f'(x):=\sum_{n=1}^{\infty} H_n\;x^n=-\frac 1{1-x}\ln(1-x)$$ (expand $\frac 1{1-x}=1+x+x^2+\cdots\ $ and $\ -\ln(1-x)=x+\frac {x^2}2+\frac{x^3}3+\cdots$ and multiply !)

Integrating $(2)$ we get : $$\tag{3}f(x)=\sum_{n=1}^{\infty} H_n \frac{x^{n+1}}{n+1}=-\int\frac{\ln(1-x)}{1-x} dx=\frac 12(\ln(1-x))^2$$ so that (using $\;H_n=H_{n+1}-\frac 1{n+1}\;$ and setting $\,k:=n+1\;$) : $$\sum_{k=2}^{\infty} H_k \frac{x^k}k-\sum_{k=2}^{\infty} \frac{x^k}{k^2}=\sum_{k=1}^{\infty} H_k \frac{x^k}k-\sum_{k=1}^{\infty} \frac{x^k}{k^2}=\frac 12(\ln(1-x))^2$$ $\operatorname{Li}_2(x):=\sum_{k=1}^{\infty} \frac{x^k}{k^2}$ is the dilogarithm but we prefer to use directly : $$\tag{4}\sum_{k=1}^{\infty} H_k \frac{x^k}k=\sum_{k=1}^{\infty} \frac{x^k}{k^2}+\frac 12(\ln(1-x))^2$$ For $x=e^i$ in $(1)$ we get following expression of $S$ : \begin{align} S&=\Im\left(\sum_{k=1}^{\infty} H_k \frac{e^{ik}}k\right)\\ S&=\sum_{k=1}^{\infty} \frac{\sin(k)}{k^2}+\frac 12\Im\left(\ln\bigl(1-e^i\bigr)\right)^2\\ S&=\boxed{\displaystyle\operatorname{Cl}_2(1)-\ln\left(2\sin\left(\frac 12\right)\right)\frac{\pi-1}2}\\ S&\approx 1.05895346485231034922735 \end{align} with $\operatorname{Cl}_2$ the Clausen function : $\displaystyle\operatorname{Cl}_2(x)=-\int_0^x \ln\left(2\sin\left(\frac t2\right)\right)dt$

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In fact, Mathematica computes the sum in closed form (well, in terms of the PolyLog anyway). Here's the computation, beginning with a numerical computation for comparison.

enter image description here

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As pointed out in the comments $\sum \frac{\sin (n)}{n}$ is a convergent series and can be read about here. Do note however that the series is not absolutely convergent - if you're working from Mattuck's book then he asks to prove that it's not absolutely convergent, not that it's not convergent at all.

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