# Sum to n terms the series $\cos \theta+ 2\cos 2\theta+ \cdots + n\cos n\theta$

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 $S_n$.

• $$\sum_{r=1}^nr\cos(rt)=$$ real part of $$\sum_{r=1}^nr(e^{it})^r$$ Aug 27 '18 at 9:01
• Relevant (uses complex exponentials and differentiation and gives the answers for sin and cos): youtu.be/X9J3Cq2_5hA Aug 27 '18 at 9:06
• alternatively, look at $\sum_{r=1}^{r=n} \sin (r \theta)$ and then take $\frac{d}{d \theta}$
– jim
Aug 27 '18 at 9:09
• If you're given the answer, you could also prove it by induction.
– J.G.
Aug 27 '18 at 9:36

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.

• $sin\theta +sin2\theta +sin3\theta... +sin n\theta=\frac{sin\frac{ n\theta}{2}}{sin\frac{\theta}{2}}.sin(n+1)\frac{\theta}{2}$ 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 Aug 27 '18 at 10:43
• I think you are right. Aug 27 '18 at 12:05
• 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 $\sin nx$. Aug 27 '18 at 13:40
• There is also a prof in this site. just write the formula and search Aug 27 '18 at 14:03
• @sirous Yeah, I have done exactly in that way, it seems very complicated. Sep 1 '18 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.

• tarit goswami, OP is clearly in precalculus. what is OP to do with this complex analysis answer? :|
– BCLC
Sep 1 '18 at 6:22
• @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. Sep 1 '18 at 6:41