I have found the following formula for the sum of cosines in both  and .  \begin{align} \sum^n_{l=1} \cos \left(\frac{2 \pi l}{n}\right) = 0 \end{align}  I would like to know what the sum would be if there is a multiplicative factor $k$ in the angle.  \begin{align} \sum^n_{l=1} \cos \left(\frac{2 \pi l k}{n}\right) = ? \end{align}  where, $k$ is a positive integer and $1 \le k \le (n-2)/2 $.  Moreover, what if the interval for $l$ is changed from $[1, n]$ to $[0, l-1]$. So,   \begin{align} \sum^{n-1}_{l=0} \cos \left(\frac{2 \pi l k}{n}\right) = ? \end{align}
1806
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:  $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$  There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$   How do we prove the above two identities?
46049
I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:  $$\begin{align}  &amp;\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} &amp; &amp;\text{(1)} \\ &amp;k2^{-k-1} = -\frac{d}{dk}2^{-k} &amp; &amp;\text{(2)} \end{align}$$   Plugging $(2)$ to $(1)$,   $$\begin{aligned}  2^{n+1}\sum_{k=1}^n k2^{-k-1}  &amp;= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\ &amp;= -2^{n+1}\cdot 0 \\ &amp;= 0 \end{aligned}$$   Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks!
86924
Find the sum of the finite series $$\sum _{k=1}^{k=89} \frac{1}{\sin(k^{\circ})\sin((k+1)^{\circ})}$$ This problem was asked in a test in my school. The answer seems to be $\dfrac{\cos1^{\circ}}{\sin^21^{\circ}}$ but I do not know how. I have tried reducing it using sum to product formulae and found out the actual value and it agrees well. Haven't been successful in telescoping it.
68251
How would I find the closed form of  $$ f(\alpha) = \sum_{n=0}^{\infty} \frac{\cos(\alpha \, n)}{n!}$$  My old pal Wolfram tells me that  $$ f(\alpha) = \mathrm{e}^{\cos(\alpha)}\cos(\sin(\alpha))$$  I've attempted writing out the first couple terms for different values of $\alpha$, but I haven't quite figured out how to arrive at the result.
112124
The series is: $$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$   Considering it isn't always positive, I replace $\frac{\cos{\sqrt{n}}}{n}$ with its absolute value and I find that: $$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac{\cos{2\sqrt{n}}}{2n}$$ if $\sum_{n=1}^\infty \vert\frac{\cos{\sqrt{n}}}{n}\vert $   converges, then  $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that  $\sum_{n=1}^\infty\frac{1}{2n}$  converges , which is impossible. So I get that $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$  absolutely diverges. But I can't figure out whether  $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges or not.  I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=\sum_{k=1}^n \cos{\sqrt{k}}$$ is bounded.  (This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)
105538
$$\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$$
51538
So I have this problem for midterm reviews:  $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$  I know that you can find the series form of a natural log, as shown here:  $$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$  But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.
21239
Does the series $\sum \limits _{n=0}^{\infty} \cos(n\pi)$ converge or diverge?   On substituting values I get alternate $1$ and $-1$.    So taking sum of infinite GP, I get $\dfrac 1 2$. So it looks like it converges.
16343
Evaluate the limit $$\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{[n/2]}\cos \left(\frac{k\pi}{n}\right)$$  I tried through considering two cases : (i) When $n$ is even (ii) when $n$ is odd.  When $n$ is even then the limiting value is $\frac{1}{\pi}$. But when $n$ is odd then ??  Case 1 : $n$ is even..  Then , take $n=2p$ where $p$ is positive integer. Then $$\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{[n/2]}\cos \left(\frac{k\pi}{n}\right)$$  $$=\lim_{p\to \infty}\frac{1}{2p}\sum_{k=0}^{p}\cos\left(\frac{k\pi}{2p}\right)$$  $$=\frac{1}{2}\int_0^1\cos(\pi x/2)\,dx$$  $$=\frac{1}{\pi}.$$  Case 2 : When $n$ is odd.  Take, $n=2p+1$ , where $p$ is a positive integer. Then ,   $$\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{[n/2]}\cos \left(\frac{k\pi}{n}\right)$$  $$=\lim_{p\to \infty}\frac{1}{2p+1}\sum_{k=0}^p\cos\left(\frac{k\pi}{2p+1}\right)$$  From here how I proceed ?
60097
I'm a bit confused as to this problem:  Consider the infinite series:  $$\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) \cdots$$  a) Find the sum $S_n$ of the first $n$ terms.  b) Find the sum of this infinite series.  I can't get past part a) - or rather I should say I'm not sure how to do it anymore.  Because the problem asks for a sum of the infinite series, I'm assuming the series must be Geometric, so I tried to find the common ratio based on the formula  $$r = S_n / S_{n-1} = \frac{\frac 12 - \frac 13}{1 - \frac 12} = \frac 13$$  Which is fine, but when I check the ratio of the 3rd and the 2nd terms:  $$r = \frac{\frac 13 - \frac 14}{\frac 12 - \frac 13} =\frac 12$$  So the ratio isn't constant... I tried finding a common difference instead, but the difference between 2 consecutive terms wasn't constant either.  I feel like I must be doing something wrong or otherwise missing something, because looking at the problem, I notice that the terms given have a pattern:  $$\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) + \cdots + \left(\frac 1n - \frac{1}{n+1}\right)$$  Which is reminiscent of the textbook's proof of the equation that yields the sum of the first $n$ terms of a geometric sequence, where every term besides $a_1$ and $a_n$ cancel out and yield  $$a_1 \times \frac{1 - r^n}{1 - r}.$$  But I don't know really know how to proceed at this point, since I can't find a common ratio or difference.
122128
Prove the following:  $\displaystyle\prod_{n=1}^{180}\left(\cos{\left(\dfrac{n\pi}{180}\right)}+2\right)=\displaystyle\sum_{n=0}^{89}\binom{180}{2n+1}\left(\dfrac{3}{4}\right)^n$.  We can state this as $\displaystyle \binom{180}{1}\left(\dfrac{3}{4}\right)^0+ \binom{180}{3}\left(\dfrac{3}{4}\right)^1+\cdots+\binom{180}{179}\left(\dfrac{3}{4}\right)^{89} = \left(\cos\left(\dfrac{\pi}{180}\right)+2\right)\left(\cos\left(\dfrac{2\pi}{180}\right)+2\right) \cdots \left(\cos\left(\dfrac{180\pi}{180}\right)+2\right)$   The problem I am encountering is that there is no general result we can prove about this for $n$ it seems. I tried using induction on $n$ to prove this but it fails for the base case. Also, combinations and cosines seem unrelated, so how could I prove the equation is true?
58007
I found the following formula   $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$  and it is cited that Euler proved the formula above , but how ?  Do there exist other proofs ?  Can we have a general formula for the alternating form   $$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$
39887
This is a homework question whereby I am supposed to evaluate:  $$\sum_{n=1}^\infty \frac{1}{n^2 +1}$$  Wolfram Alpha outputs the answer as   $$\frac{1}{2}(\pi \coth(\pi) - 1)$$  But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
97715
My Prof wrote that   $\sum_{n=2}^{\infty} n(n-1)x^{(n-2)} = \frac{2}{(1-x)^3}$  when $-1&lt; x&lt;1$  I do not understand where the above formula comes from. Can someone please explain the steps or provide a hint?   Thanks!
89519
So I have been trying for a few days to figure out the sum of   $$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical analysis and CAS that this sum equals  $$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$  But I have not been able to prove this yet. Anyone know how? My guess is that the  sum of this series is related to fourier-series but nothing particalr comes to mind.   For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.
75611
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$   MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$ = Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$  $\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$= $\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$  I don't think i can telescope it, And i don't know any other method
76802
The following fascinating formula appears in the paper  by M.Chamberland and A.Straub (see page 9): $$\prod_{n=2}^\infty\left(1-\frac{1}{n(n-1)}\right)=-\frac1\pi\cos\left(\frac{\sqrt5}{2}\pi\right).$$ Unfortunately, it's given without any reference or explanation. Could anyone give one?