# Math Help - Sum to infinity of series

1. ## Sum to infinity of series

Show that $n(n+1)\equiv(n-1)(n-2)+4(n-1)+2$ and hence, that for $n\geq 3$,
$\frac{n(n+1)}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!}$
Deduce or find otherwise the sum to infinity of the series
$1.2+\frac{2.3}{1!}+\frac{3.4}{2!}+...+\frac{n(n+1) }{(n-1)!}+...$

I've done the first two parts correctly, but I don't know how to do this last part.
$\sum^{n}_{r=1} \frac{r(r+1)}{(r-1)!}=\sum^{n}_{r=1}\frac{1}{(r-3)!}+\frac{4}{(r-2)!}+\frac{2}{(r-1)!}$
This is the furthest I got, where $n<3$, I can't find values, so I don't know how else to find the sum to n, or to infinity.
Thanks!

2. Originally Posted by arze
Show that $n(n+1)\equiv(n-1)(n-2)+4(n-1)+2$ and hence, that for $n\geq 3$,
$\frac{n(n+1)}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!}$
Deduce or find otherwise the sum to infinity of the series
$1.2+\frac{2.3}{1!}+\frac{3.4}{2!}+...+\frac{n(n+1) }{(n-1)!}+...$

I've done the first two parts correctly, but I don't know how to do this last part.
$\sum^{n}_{r=1} \frac{r(r+1)}{(r-1)!}=\sum^{n}_{r=1}\frac{1}{(r-3)!}+\frac{4}{(r-2)!}+\frac{2}{(r-1)!}$
This is the furthest I got, where $n<3$, I can't find values, so I don't know how else to find the sum to n, or to infinity.
Thanks!
$\displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum_{n=3}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum^{\infty}_{n=3}\left(\frac1{(n-3)!}+\frac4{(n-2)!}+\frac2{(n-1)!}\right)$

Now use the fact that $\displaystyle e=\sum_{n=0}^\infty \frac1{n!}$

3. Originally Posted by chiph588@
$\displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum_{n=3}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum^{\infty}_{n=3}\left(\frac1{(n-3)!}+\frac4{(n-2)!}+\frac2{(n-1)!}\right)$

Now use the fact that $\displaystyle e=\sum_{n=0}^\infty \frac1{n!}$
so i have
$\displaystyle 8+(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!}-\frac{1}{(n-1)!}-\frac{1}{(n-2)!})+2(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!}-\frac{1}{(n-1)!})+4(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!})$
$8+(e-1-0-0-0)+2(e-1-0-0)+4(e-1-0)=7e+1$
Answer is 7e, the eight is the problem it think
Thanks

4. $\displaystyle \sum_{n=3}^\infty \frac1{(n-3)!} = \sum_{n=0}^\infty \frac1{n!} = e$

$\displaystyle \sum_{n=3}^\infty \frac4{(n-2)!} = 4\sum_{n=1}^\infty \frac1{n!} = 4\sum_{n=0}^\infty \frac1{n!}-4 = 4e-4$

$\displaystyle \sum_{n=3}^\infty \frac2{(n-1)!} = 2\sum_{n=2}^\infty \frac1{n!} = 2\sum_{n=0}^\infty \frac1{n!}-2-2 = 2e-4$