Find the sum of the first $50$ terms of the series $$\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....$$   $$ \sum_1^{50}=\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21+.....\\ =\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}+.....= $$ My reference gives the solution $\tan^{-1}\dfrac{5}{6}$, but I do not have any clue of doing it ?  Note: I know that $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ if $xy&lt;1$.
4261
This is from a GRE prep book, so I know the solution and process but I thought it was an interesting question: Explicitly evaluate $$\sum_{n=1}^{m}\arctan\left({\frac{1}{{n^2+n+1}}}\right).$$
49569
How to find the sum of this series?  $$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$  It seems very easy. But I still can not work it out, can anyone help?
39958
I have this problem :  $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$  My solution  $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$  I think that the sum should be $\frac{1}{2}$ since the limit of :  $$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$  But that wrong. Any ideas?
80108
I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it. $$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$  Thanks.
30027
A series is given as follows   $$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$  Can you give me hints to get started finding its value? Thanks.
38743
I'm in trouble with this homework.  Find the sum of the series $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$, where the terms are the inverse of the positive integers whose only prime factors are 2 and 3.  Hint: write the series as a product of two geometric series.  Ok, I found some patterns in these numbers, but I can't find the two series.
64473
$$\tan A + \tan 2A + \tan 3A=0$$ I tried converting these all in sin and cos and I got the answer but the answer didn't match.  So any one could just help me.  Thanks
97514
I am trying to sum the following series, given n is a positive integer, $$ n+n(n-1)+n(n-1)(n-2)+...+n! $$ I think there is a solution, but if not, an approximate result will also be appreciated.   Regards.
36528
How to find the sum of the following series:  $\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$  Any hints.
116013
I have to prove the following result :  $$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$  I tried converting $\tan\theta$ &amp; $\cot\theta$ into $\cos\theta$ and $\sin\theta$. That led to a huge expression which I wasn't able to simplify.  Please help!!!
107955
I found a nice general formula for the Taylor series of $\tan x$:  $$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$  where $B_n$ are the Bernoulli numbers and $|x| &lt; \dfrac {\pi} 2$.  I've tried Googling for a proof but didn't find anything. Hints would be appreciated too.  I am using the typical definition of the Bernoulli numbers:  $$\frac x {e^x - 1} = \sum_{n\,=\,0}^\infty \frac {B_n x^n} {n!}$$
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.
6112
Find the sum $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2}(2n-1)}$  $= 1 + \dfrac{1}{3\cdot 2^{2}} + \dfrac{1}{5\cdot 3^{2}} + \dfrac{1}{7\cdot 4^{2}} \ldots$  Then, I'm stuck. I didn't find any pattern to solve.  I just need a hint.
12554
Does anyone know the sum of all triangle numbers? I.e  1+3+6+10+15+21...  I've tried everything, but it might help you if I tell you one useful discovery I've made:   I know that the sum of alternating triangle numbers, 1-3+6-10... Is equal to 1/8 and that to change 1+3+6... Into 1-3+6... You would subtract 6+20+42+70... which is every other triangular number (not the hexagonals) multiplied by two.   1/8 plus this value is 1+3+6+10+...  A final note: I tried to split the triangle numbers into hexagonals and that series and then I got the squares of the odd numbers. Using dirichlet lambda functions This gave me 0 but I don't think this could be right. A number of other sums gave me -1/24 and 3/8 but I have no idea
110019
Find the sum of this series.   I tried writing the general term of this term. I succeeded in it but the sum inside the bracket depends on r which makes more problem to get overall sum.  I know that this uses the concept of  binomial theorem and binomial coefficients but I am not able to break through it to get the desired concept to use.
69384
By looking through an book, I found this interesting series  To prove that:  $$\tan(\theta)+\tan \left(\theta+ \frac{\pi}{n} \right) + \tan\left(\theta + \frac{2\pi}{n}\right) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$  I have tried the Gaussian pairing trick and using:  $$\tan A + \tan B = \tan(A+B) (1-\tan A \tan B)$$  And  $$\tan A \tan B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B)+\cos(A+B)}$$  However I could not do anything of great use.  I also considered perhaps roots of unity of:  $$z^{2n} = \cos \theta + i\sin \theta$$  But that grew to no real use, but it may be an idea for others.  I know I can sum an arithmetic series of the arguments for the cosine and sine functions but I cannot find a source on the internet about the tangent function.   Thank You.
3137
If in the series  $1-\dfrac {1} {2}+\dfrac {1} {3}-\dfrac {1} {4}+\ldots $ the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first $n$ terms is ultimately $a^{2}$, show that the sum of the series will become $\log \left( 2a\right) $.  Solution attempt. Let $p$ be the number of positive terms in the first $n$ terms of the reordered series so based on the question we are allowed $a^{2}=\dfrac {p} {n-p}$. Now solving for p we get $p=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }$ and $n-p=\dfrac {n } {\left( 1+a^{2}\right) }$.  We also observe that in the original series only the odd terms are positive and only the even terms are negative.  Let's define $S_{odd}=1+\dfrac {1} {3}+\dfrac {1} {5}+.\ldots +\dfrac {1} {2n-1}$ and $S_{even}=\dfrac {-1} {2}\dfrac {-1} {4}\ldots -\dfrac {1} {2n}$  $S_{Reordered}=S_{odd_{p}}+S_{even_{n-p}}$  $S_{Reordered}=\sum _{t=1}^{t=P}\dfrac {1} {2t-1}+\sum _{t=1}^{t=(n-P)}\dfrac {-1} {2t}$ $S_{Reordered}=\sum _{t=1}^{t=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\dfrac {n } {\left( 1+a^{2}\right) }}\dfrac {-1} {2t}$    In order to extend $S_{Reordered}$ from n terms to an infinite length. I guess i should take the limit of the $S_{Reordered}$ as $n\rightarrow \infty $ which would make the upper values of $t$ (Not sure of technical term here) to be $\infty$ giving me   $S_{Reordered}=\sum _{t=1}^{t=\infty}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\infty}\dfrac {-1} {2t}$   and i have lost the $a$ from the expression. I guess i am stuck i need to perform some step to capture a before taking the limit. Any help would be much appreciated.