I learned that $$\sum \limits_{i=1}^n i(i+1) = \frac{n(n+1)(n+2)}{3}$$ or in general $$\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \frac{n(n+1)\dots (n+k+1)}{k+2}$$ From a mathematical standpoint why is this true? I'm not asking for inductive proof. I am asking if you only given the left hand side, how would you go about writing a closed form expression for the sum?
35691
Considering the following formulae:  (i) $1+2+3+..+n = n(n+1)/2$  (ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$  (iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$  Find and prove a 'closed formula' for the sum  $1\cdot2\cdot3\cdot...\cdot k + 2\cdot3\cdot4\cdot...\cdot(k+1) + ... + n(n+1)(n+2)\cdot...\cdot (k+n-1)$  generalizing the formulae above.  I have attempted to 'put' the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.
1154
I am trying to compile a list of the different ways to come up with the closed form for the sum in the title. So far I have the famous Gauss story, the argument by counting doubletons, and using generating functions. Is there any other (significantly) different way to discover the closed form?  To clarify, Gauss's method was to take $1+2+3+\dots+n$, write it backwards, and add it up.
63339
What is $\sum\limits_{i=1}^n\sqrt i\ $?   Also I noticed that $\sum\limits_{i=1}^ni^k=P(n)$ where $k$ is a natural number and $P$ is a polynomial of degree $k+1$. Does that also hold for any real positive number? How could one prove it?
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?
25555
I'm finding a closed form to  $\sum^{n}_{k=1} \frac{k}{(k+1)!}  (n \leq 1) $ (in a environment of induction and recurrence)   I've been trying to solve it without success, can anybody help me (?)  The last thing I've tried to do is write the expression as a Non-Homogeneous Linear Recurrence:   $ f(1) = \frac{1}{2} $   but, actually I have no idea of how to attach it.   $ f(n+1) = f(n) + \frac{1}{(n+2)n!} $
38461
I am trying to calculate the following series:  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$  and I managed to reduce it to this term  $$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$  And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$  How should I go about solving this question
28782
I am trying to obtain a closed form expression for  $\sum {n \choose 4k}$  I am trying to use the binomial expansion of $(1 + i)^n$ and $(1 - i)^n$. $$(1 + i)^n + (1 - i)^n = 2\left(\sum {n \choose 4k} -\sum {n \choose 4k + 2}\right)$$ Stuck at this now. I can't come up with a way to simplify the second term on RHS. Any help would be appreciated.
75725
Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$  I tried using integration but failed miserably. Hints please.
115435
I know that  $$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$ and  $$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$ Here is the question: is there a formula for  $$\sum^n_{i=1}i^4.$$
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Mathematica tells me that  $$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$  Although I have not been able to come up with a proof.  Proofs, hints, or references are all welcome.
94106
Let $N$ be a positive integer. Then, we have  $\sum\limits_{j=1}^{N} \binom{j}{6} = \binom{N+1}{7}$.  Could anyone explain this equation a little bit? I wrote out the left hand side as $\binom{1}{6} + \dots + \binom{N}{6}$, but it did not help at much.  UPDATE  From lecture notes, I found a proposition says:     $\sum\limits_{0\leq k\leq n} \binom{k}{m} = \binom{n+1}{m+1}$.   Could anyone give a hint on how to prove it?
2301
I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done?
77605
A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.
11067
Is my approach to evaluating this series correct?   $$\sum_{n=1}^\infty \frac{n}{(n+1)!}$$  Has partial sum equivalent to:  $$S_m = \sum_{n=1}^m \frac{n}{(n+1)!} = \sum_{j=2}^{m+1} \frac{j-1}{j!} = \sum_{j=2}^{m+1} \frac{1}{(j-1)!} - \sum_{j=2}^{m+1} \frac{1}{j!} $$  For $j$ such that $m+1&gt;j&gt;2$ the terms of the left sum are cancelled by the terms of the right, leaving  $$ S_m =1-\frac{1}{(m+1)!}$$  Hence $ \lim_{m\rightarrow\infty} S_m  = 1$  Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!
30466
I found the following identity when trying to prove something else: $$\sum\limits_{n=1}^\infty\frac{(n-1)!x}{\prod\limits_{k=1}^n (k+x)}\equiv 1,~ \forall x&gt;0$$ I'm sure there's a name for it (or for a more general identity for which this is a special case). Does anyone know the name?
34046
I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$  It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?