1. ## sum

evaluate the series $\sum_{k\,=\,2}^{\infty }{\frac{1}{(k+1)(k+2)(k+3)}}$ as many ways as possible

2. Originally Posted by Ajreb
evaluate the series $\sum_{k\,=\,2}^{\infty }{\frac{1}{(k+1)(k+2)(k+3)}}$ as many ways as possible
$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

So $\iiint\frac{dx}{1-x}\bigg|_{x=1}=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n +2)(n+3)}$

EDIT: Sorry, now I have time to finish.

So we know that \begin{aligned}\iiint\frac{dx}{1-x}&=\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\\
&=\iiint\sum_{n=0}^{\infty}x^n\\
&=\sum_{n=0}^{\infty}\frac{x^{n+3}}{(n+1)(n+2)(n+3 )}~\forall{x}\in[-1,1]
\end{aligned}

So

\begin{aligned}\bigg[\frac{-1}{2}\ln(1-x^2)(1-x)^2+\frac{1}{4}-\frac{3}{2}x+\frac{3}{4}x^2\bigg]\bigg|_{x=1}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+ 2)(n+3)}\\
&=\frac{1}{4}
\end{aligned}

So finally \begin{aligned}\sum_{n=2}^{\infty}\frac{1}{(n+1)(n +2)(n+3)}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)( n+3)}-\frac{1}{6}-\frac{1}{24}\\
\end{aligned}

3. Here's my approach:

Your series equals $\int_{0}^{1}{\int_{0}^{1}{\left( y\sum\limits_{k=2}^{\infty }{\frac{(xy)^{k+1}}{k+1}} \right)\,dx}\,dy},$ and after some simple calculations this is $-\left( \int_{0}^{1}{\frac{y^{3}+3y^{2}-6y}{6}\,dy}+\int_{0}^{1}{(y-1)\ln (1-y)\,dy} \right).$ First integral is easy to solve, so for the second one I'll perform a double integration solution, hence the series equals $\frac{7}{24}+\int_{0}^{1}{\int_{0}^{y}{\frac{y-1}{1-z}\,dz}\,dy}=\frac{7}{24}-\frac{1}{4}=\frac{1}{24}.$

4. Ill just go ahead and be the one to suggest the ugliness that is PFD followed by Telescoping Series for this.

$\sum_{n=2}^{\infty}\left\{\frac{1}{2(n+3)}-\frac{1}{n+2}+\frac{1}{2(n+1)}\right\}$

If anyone would like to do it I would be interested to see how many terms you would have to go through to see the actual sum show its face.

5. both methods given by Mathstud28 and Krizalid are good but they're not strong enough to handle the general form, i.e. evaluating: $\sigma_n=\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)},$ where $n \geq 2$ is given.

note that the sum here starts from $k = 0,$ which is more usual than k = 2. evaluating $\sigma_n$ is quite easy if you're familiar with gamma and beta functions. here's the solution: first observe that:

$\frac{1}{(k+1)(k+2) \cdots (k+n)}=\frac{k!(n-1)!}{(n-1)!(k+n)!}=\frac{\Gamma(k+1)\Gamma(n)}{(n-1)!\Gamma(k+n+1)}=\frac{B(k+1,n)}{(n-1)!}.$ from here things are straightforward:

$(n-1)!\sigma_n=\sum_{k=0}^{\infty}B(k+1,n)=\sum_{k=0} ^{\infty}\int_0^1t^k(1-t)^{n-1} \ dt=\int_0^1[(1-t)^{n-1}\sum_{k=0}^{\infty}t^k] \ dt$

$=\int_0^1\frac{(1-t)^{n-1}}{1-t} \ dt = \int_0^1 (1-t)^{n-2} \ dt = \frac{1}{n-1} \Longrightarrow \sigma_n=\frac{1}{(n-1)(n-1)!}. \ \ \Box$

as a result, if you really want your sum to start from $k = 2,$ you will get: $\sum_{k=2}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}=\sigma_n - \frac{1}{n!} - \frac{1}{(n+1)!}=\frac{2}{(n-1)(n+1)!}.$

6. Originally Posted by NonCommAlg
both methods given by Mathstud28 and Krizalid are good but they're not strong enough to handle the general form, i.e. evaluating: $\sigma_n=\sum_{k=0}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)},$ where $n \geq 2$ is given.

note that the sum here starts from $k = 0,$ which is more usual than k = 2. evaluating $\sigma_n$ is quite easy if you're familiar with gamma and beta functions. here's the solution: first observe that:

$\frac{1}{(k+1)(k+2) \cdots (k+n)}=\frac{k!(n-1)!}{(n-1)!(k+n)!}=\frac{\Gamma(k+1)\Gamma(n)}{(n-1)!\Gamma(k+n+1)}=\frac{B(k+1,n)}{(n-1)!}.$ from here things are straightforward:

$(n-1)!\sigma_n=\sum_{k=0}^{\infty}B(k+1,n)=\sum_{k=0} ^{\infty}\int_0^1t^k(1-t)^{n-1} \ dt=\int_0^1[(1-t)^{n-1}\sum_{k=0}^{\infty}t^k] \ dt$

$=\int_0^1\frac{(1-t)^{n-1}}{1-t} \ dt = \int_0^1 (1-t)^{n-2} \ dt = \frac{1}{n-1} \Longrightarrow \sigma_n=\frac{1}{(n-1)(n-1)!}. \ \ \Box$

as a result, if you really want your sum to start from $k = 2,$ you will get: $\sum_{k=2}^{\infty} \frac{1}{(k+1)(k+2) \cdots (k+n)}=\sigma_n - \frac{1}{n!} - \frac{1}{(n+1)!}=\frac{2}{(n-1)(n+1)!}.$
Hmm, this is cool because melding my method and yours together

$\underbrace{\int_0^1\int\cdots\int\frac{dx}{1-x}}_{n\text{ times}}=\frac{1}{(n-1)(n-1)!}$

7. Originally Posted by Mathstud28
Hmm, this is cool because melding my method and yours together

$\underbrace{\int_0^1\int\cdots\int\frac{dx}{1-x}}_{n\text{ times}}=\frac{1}{(n-1)(n-1)!}$
right! you can rewrite it more clearly: $\int_0^1\int_0^{x_n} \cdots \int_0^{x_3} \int_0^{x_2} \frac{\ dx_1 \ dx_2 \ \cdots \ dx_n}{1-x_1} =\frac{1}{(n-1)(n-1)!}, \ \ 0 < x_j < 1.$

the challenge now is to prove this result directly, i.e. without using the series!

8. Originally Posted by NonCommAlg
right! you can rewrite it more clearly: $\underbrace{\int_0^1\int_0^{x_n} \cdots \int_0^{x_3} \int_0^{x_2}}_{n\text{ times}} \frac{\ dx_1 \ dx_2 \ \cdots \ dx_n}{1-x_1} =\frac{1}{(n-1)(n-1)!}, \ \ 0 < x_j < 1.$

the challenge now is to prove this result directly, i.e. without using the series!
I will come back tomorrow and give it a go, but right now I am thinking induction. And maybe playing around with some mroe generalized functions so that I can use Leibniz's Rule of Differentiation Under the Integral Sign in the inductive step.