The problem is to determine the sum $$\sum_{n=1}^\infty \frac 1{2^n} \tan \frac 1{2^{n+1}}$$  I have been trying everything that comes to my mind (trigonometrical functions' formulas, transforming to integration, Taylor expansion, etc.) to no avail, so a slightest hint to a successful approach would be much appreciated.
122589
Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$   I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into $$\lim_{n\to \infty} \sum_{r=1}^n \left(\frac {1}{2^r(1-\tan (2^{-r+1})} -\frac {1}{2^r(1+\tan (2^{-r+1})}\right) $$  But couldn't proceed further. Also I thought about Riemann sums but it was a pure dead end.  Any help would be greatly appreciated
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.
46818
Find the following sum : $\sum_{n=1}^{\infty} \frac{\ln n }{n(n- \ln n)}$  I couldn't find any way to solve.   I just want a hint, how to initiate?
122835
I found the following formula here: .  Taylor series of $\tan x$:  $$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$.  I do not understand what the Bernoulli number means in this formula. My problems are the following ones:  I do not know anything about Bernoulli numbers. I do not know how to expand the Bernoulli number in the formula you see further up. What expansion will be the final result? How did you arrive there? Why did you arrive there? How did you know that you should use the specific Bernoulli-related terms that you used in your answer? What is the purpose of the Bernoulli number in this formula?  A great deal of thanks.
17273
I was asked to evaluate the following sum: $$\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}dx$$  I'm trying to use $$\int_{0}^{\infty} \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ However, it doesn't seem to work. Any help is greatly appreciated.
89659
I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$  my steps:   $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum$$  I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..
9714
Find this sum $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}$$  My try: since $$n^4+n^2+1=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$ then $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}=\sum_{n=1}^{\infty}\dfrac{1}{2n\cdot n!}\left(\dfrac{1}{n(n-1)+1}-\dfrac{1}{n(n+1)+1}\right)$$ then I can't go further.
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
54251
Show that the following sum is divergent   $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$$   I thought to do this using Taylor series using the fact that $$ \ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right) $$ Which then makes it clear that $$ \sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty $$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?
9386
Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$  I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
53324
How can I prove that $$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$  I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
63488
Evaluate the sum  $$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$  Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:  $$\sum_{k=0}^{\infty}\frac{(2k)!!}{(2k+1)!!}x^{2k+1}.$$
103332
Can anyone help me with this?  What is the sum of this series:  $\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$  I got it after plugging $x=-1$ in a Fourier series   Thank you!
13007
What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you!
48257
Show that: $$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$ $$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$ Thank you in advance. NOTE: The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$, which is a mistake as noted by the comments below.
64621
Can anyone tell me about the sum of the series $$\sum_{n=1}^\infty \frac{1}{(2n)(2n+1)}?$$ This is not a usual telescoping sum in which all the terms cancel out.
107744
Here is the series:  $$ \sum_{k = 0}^{\infty} \frac{(-1)^k}{2k + 1}. $$  I don't know how to start at all. Thank you for your help.