$$\sum_{k=1}^{519}\frac{1}{k(k+2)}$$  I've been trying to figure out this summation for a while now but I can't seem to get it. I've been looking at my notes and the only trick I've been given for summations like this one is $\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$. This doesn't help me to my knowledge because this kind of fraction decomposition doesn't work on my question. I have premium Symbolab and that says it doesn't support this kind of question. WolframAlpha gives me the decimal approximation of 0.7480 but no support work.  I'm sorry if this question has been posted here before but I couldn't find this question anywhere online.  Thank you for the help,  -jjleahy
120733
More specifically, I'm supposed to compute $\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} $ by using the equality $\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}$ and the problem before which just says that, $\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0$.  I can add up the sum for any $n$ but I'm not sure what they mean by "compute".   Thanks!
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..
97681
I'm trying to prove that  $$\sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1} = \frac{1}{n+1}$$  So far I've tried induction (which doesn't really work at all), using well known facts such as  $$\sum_{k=0}^n {n \choose k} (-1)^k = 0$$  and trying to apply identities like  $${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$  Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing?
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
117844
I am wondering how to evaluate the following sum:  $$\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}.$$  In wolfram alpha I find it is equal to $e/2$ .  I have used the residue method but I didn't succeed and also using digamma function is still hard for me, my problem is treating the $n!$.
8533
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
4063
I've been trying to prove   $$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$  I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand . Is there any proof that doesn't require the integral transform?
123432
I have the following series  $$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$  I have found out that the series is equal to  $$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$  I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.  Does anyone have any idea how I can go about doing this ?
23460
How to find the sum of the following series? $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$ This is a harmonic progression. So, is the following formula correct? $\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$ $\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P. Therefore the sum of the series can be written as : $\Rightarrow \frac{(3)^3}{(A+B+C)}$ Is this correct? Please suggest.
53213
i know the summation of $1 \times 2 + 2\times 3 + 3\times 4 = \frac{k(k+1)(k+2)}3$ but this one has negative terms. so we need to find below  $$\sum_{i=1}^n(-1)^{i+1} (i(i+1))$$  e.g. $1\times 2 - 2\times 3 + 3\times 4 ....$  if you can share the derivation then it would be very great.
64081
I have to show that the double serie $$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} ((\frac{1}{k+1} \cdot (\frac{k}{k+1})^{i}) - (\frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}))$$ convergence to $-\frac{1}{2}$. Wolfram Alpha also says this series convergence to $-\frac{1}{2}$. Unfortunately, I don't know the calculation route.
110400
I need to prove the following: $$\sum_{k=0}^n {n \choose k}\cdot\frac{(-1)^k}{k+1} = \frac{1}{n+1}$$  Started from the left, I got this far:  $$\frac{1}{n+1}+\sum_{k=1}^n {n+1 \choose k+1}\cdot(-1)^k$$  Am I right so far? Any ideas how to move on? thanks.  Edit: I guess my last move had a mistake, then perhaps this is ok? $$\frac{1}{n+1}\cdot\sum_{k=0}^n {n+1 \choose k+1}\cdot(-1)^k$$  How do you suggest I move on?
25201
My partner is tutoring a Civil Engineering student in maths. Conveniently, I am a civil engineer so when any maths that's confusing comes up, I can usually help out. However, we are having a problem with the summing of power series, one area where I am sorely lacking.  After spending two hours on a question, we still cannot solve it and are always left with a $k$. Admittedly, I don't fully know what I am doing.     Evaluate $\displaystyle \sum_{k=1}^\infty\frac{x^k}{k(k+1)}$.   We do know that the answer is $1 + \dfrac{1-x}{x}\ln (1-x)$.  Any hints or help in the direction to go would be very much appreciated.
49662
How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$  What I tried as follows :   $$\sum^n_{k=1} (k^2+k+1)k!$$  =$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$  Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks.......
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.
34875
How do I calculate sum of a finite harmonic series of the following form? $$\sum_{k=a}^{b} \frac{1}{k} = \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \cdots + \frac{1}{b}$$ Is there a general formula for this? How can we approach this if not?
86635
I try to solve the following sum:  $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$  I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.