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?

  • 1
    $\begingroup$ 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

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 : $\displaystyle\operatorname{Cl}_2(x)=-\int_0^x \ln\left(2\sin\left(\frac t2\right)\right)dt$


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


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.


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.