Sum to $n$ terms and also to infinity of the following series:$$\cos \theta+ 2\cos 2\theta+ \cdots + n\cos n\theta$$the solution provided by the book is $$S_n=\frac{(n+1)\cos n\theta-n\cos(n+1)\theta-1}{2(1-\cos\theta)}$$Can anyone help me to explain how to get $S_n$.

Thanks in advance.

  • 2
    $\begingroup$ $$\sum_{r=1}^nr\cos(rt)=$$ real part of $$\sum_{r=1}^nr(e^{it})^r$$ $\endgroup$ Aug 27, 2018 at 9:01
  • $\begingroup$ Relevant (uses complex exponentials and differentiation and gives the answers for sin and cos): youtu.be/X9J3Cq2_5hA $\endgroup$ Aug 27, 2018 at 9:06
  • 1
    $\begingroup$ alternatively, look at $\sum_{r=1}^{r=n} \sin (r \theta)$ and then take $\frac{d}{d \theta}$ $\endgroup$
    – jim
    Aug 27, 2018 at 9:09
  • 1
    $\begingroup$ If you're given the answer, you could also prove it by induction. $\endgroup$
    – J.G.
    Aug 27, 2018 at 9:36

2 Answers 2


Hint: Take integral you get:

$$ (\sin x+\sin 2x+\sin 3x+\dots+\sin nx)=\frac{\cos x/2 - \cos (n+ 1/2)x}{2 \sin x/2} $$

Now take the derivative.

  • $\begingroup$ $sin\theta +sin2\theta +sin3\theta... +sin n\theta=\frac{sin\frac{ n\theta}{2}}{sin\frac{\theta}{2}}.sin(n+1)\frac{\theta}{2}$ I couldn't understand how did you get $\frac{cos\frac{x}{2}-cos(n+1/2)x}{2sin\frac{x}{2}}$.could you explain please.And thank you for your Hint.@sirous $\endgroup$
    – emonHR
    Aug 27, 2018 at 10:43
  • $\begingroup$ I think you are right. $\endgroup$
    – Narasimham
    Aug 27, 2018 at 12:05
  • 1
    $\begingroup$ md emon, it has a rather long procedure. Use expansion of $e^{i X}$ for $X= x, 2x, 3x ...$. The sum of LHS gives a geometric progression its result must be found. The RHS gives the sum of $ \cos nx$ and $\sin nx$. Writing LHS $e^{S}i=\cos s + i sin s$ gives the result where $\cos s$ is for the sum of $\cos ns$ and $\sin s$ gives the sum of $\sin nx$. $\endgroup$
    – sirous
    Aug 27, 2018 at 13:40
  • $\begingroup$ There is also a prof in this site. just write the formula and search $\endgroup$
    – sirous
    Aug 27, 2018 at 14:03
  • $\begingroup$ @sirous Yeah, I have done exactly in that way, it seems very complicated. $\endgroup$ Sep 1, 2018 at 6:47

To find closed form for series of $\cos$ and $\sin$ functions, approach using complex numbers helps very much. Consider, $z=e^{i\theta}$, where, $i=\sqrt{-1}$. We need to construct a function whose real part or imaginary part(otherwise we will have $i$ in the sum, which we don't have in the sum we are interested to find out) will give the required sum. Now, if we take $f(z)=1+z+z^2+z^3+\cdots +z^n$, $$z\cdot f'(z)=z+2z^2+3z^3+\cdots+nz^{n}\\=\cos{\theta}+2\cos{2\theta}+\cdots+\cos{n\theta}+i\sum_{k=1}^{n}k\cdot\sin{(k\theta)} $$ So, notice that $\Re{\{z\cdot f'(z)\}}$ is the required sum. On the other hand, $f(z)=\frac{z^{n+1}-1}{z-1} $ gives, $$\require{cancel}z\cdot f'(z)=z\cdot \frac{(z-1)\cdot (n+1)z^n-z^{n+1}+1}{(z-1)^2}=z\cdot \frac{(n+\cancel{1})z^{n+1}-(n+1)z^n-\cancel{z^{n+1}}+1}{(z-1)^2}\\= z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2} $$

Now, $\sum_{k=1}^{n}k\cdot \cos{(k\theta)}=\Re{\{z\cdot f'(z)\}}=\Re{\{z^{n+1}\cdot \frac{n(z-1)-1}{(z-1)^2}\}}$. After some calculation you will get the result.

  • 1
    $\begingroup$ tarit goswami, OP is clearly in precalculus. what is OP to do with this complex analysis answer? :| $\endgroup$
    – BCLC
    Sep 1, 2018 at 6:22
  • $\begingroup$ @BCLC Some introductory knowledge$\big(e^{ix}=\cos{x}+i\sin{x}\big)$ of Complex numbers will suffice to understand this :D . Though to be sure that differentiation works for complex polynomial will need complex analysis, OP can take it for granted now. This way of using $e^{ix}$ makes calculation of closed forms of lots of complicated series easier. $\endgroup$ Sep 1, 2018 at 6:41

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.