1. ## pls help

$S_1 = \sum_{k = 1}^n k cos k$

$S_2 = \sum_{k = 1}^n k sin k$

$S_1, S_2 = ? using: S_1 + iS_2$
pls help
thanx

2. Originally Posted by ely_en
$S_1 = \sum_{k = 1}^n k cos k$

$S_2 = \sum_{k = 1}^n k sin k$

$S_1, S_2 = ? using: S_1 + iS_2$
pls help
thanx
What I understand from S1 is that is is $S_1 = \sum_{k = 1}^n cos k$ = cos1+cos2+cos3+cos4+ .... +cos(n-1)+cosn which is kx{[cos(n)]x[cos(n+1)]/2}

and the same thing for S2 (Im assuming) and then adding up as in x+iy form.

Hope that helped...

3. $S_1 = \sum_{k = 1}^n k*cos k, not \sum_{k = 1}^n cos k$

4. Hello, ely_en!

I haven't solved it yet . . . but I ran across something interesting.

$S_1 \:= \:\sum_{k=1}^n k\cos k$

$S_2 \:=\: \sum_{k=1}^n k\sin k$

Find $S_1,\:S_2$ . . . using: $S_1 + iS_2$

. . . $S_1 \;=\;1\cos1 + 2\cos2 + 3\cos3 + 4\cos4 + \hdots$
. . . $iS_2 \:=\:i\sin1 + 2i\sin2 + 3i\sin3 + 4i\sin4 + \hdots$

Add: . $S_1+iS_2 \;=\;(\cos1+i\sin1) +2(\cos2 + i\sin2) + 3(\cos3 + i\sin3)$ $+ 4(\cos4+i\sin4) + \hdots$

And we have: . . . $S_1+iS_2 \;=\;e^i + 2e^{2i} + 3e^{3i} + 4e^{4i} + \hdots\;\;{\color{blue}[1]}$

Multiply by $e^i\!:\;\;e^i(S_1 + iS_2) \;=\;\qquad \;e^{2i} + 2e^{3i} + 3e^{4i} + \hdots\;\;{\color{blue}[2]}$

$\text{Subtract }{\color{blue}[1] \text{ - } [2]}\!:\;\;(1-e^i)(S_1+iS_2) \;=\;\underbrace{e^i+ e^{2i} + e^{3i} + e^{4i} + \hdots}_{\text{geometric series}}$

$\text{Hence: }\;(1-e^i)(S_1+iS_2) \;=\;\frac{e^i}{1-e^i}$

$\text{Therefore: }\;S_1+iS_2 \;=\;\frac{e^i}{(1-e^i)^2}$

A fascinating result . . .
. . absolutely useless, but still fascinating.

5. thanx
I've tried this too, but I used the moivre's formula and then the formula for the sum of a geometric progression.

6. Lemma: $\sum_{n=1}^{N}nx^n=\frac{x^{N+1}\left(xN-N-1\right)+x}{(x-1)^2}$

$\therefore\sum_{n=1}^N{n\cos(n)}$

$=\text{Re}\bigg[\sum_{n=1}^N{n\left(e^i\right)^n}\bigg]$

$=\text{Re}\bigg[\frac{\left(e^i\right)^{N+1}\left(e^iN-N-1\right)+e^i}{(e^i-1)^2}\bigg]$

$=\frac{\left((\cos(1)-1)N-1\right)\cos\left(N\right)-\sin(1)N\sin\left(N\right)+1}{2\left(\cos(1)-1\right)}$

Do the same except its

$\sum{n\sin(n)}=\text{Im}\bigg[\sum{n\left(e^i\right)^n}\bigg]$

7. Originally Posted by Soroban

$\text{Subtract }{\color{blue}[1] \text{ - } [2]}\!:\;\;(1-e^i)(S_1+iS_2) \;=\;\underbrace{e^i+ e^{2i} + e^{3i} + e^{4i} + \hdots}_{\text{geometric series}}$

$\text{Hence: }\;(1-e^i)(S_1+iS_2) \;=\;\frac{e^i}{1-e^i}$
Hmm, you seemed to apply the rule $\sum_{n=1}^{{\color{red}{\infty}}}x^n=\frac{x}{1-x}$

But this series is clearly divergent as $N\to\infty$

Besides...the OP did not ask for the infinite series.

8. Since it doesnt appear the OP isn't going to post the answer to $\sum_{n=1}^{N}n\sin(n)$ I will present it
\begin{aligned}