# series calculation involving factorial

How would one calculate following

$$\sum_{k=2}^\infty \frac{k^2+3k}{k!}$$

I searched youtube for tutorials (patricJMT and other sources) where I usually find answers for my math problems, I think I am supposed to use Differentiation but in example there is no exponent.

-
– Lucian Mar 30 '14 at 13:44

$\displaystyle\frac{k^2+3k}{k!}=\frac{k(k-1)+4k}{k!} =\frac1{(k-2)!}+4\cdot\frac1{(k-1)!}$ for $k\ge2$

So, $\displaystyle\sum_{k=2}^{\infty}\frac{k^2+3k}{k!}=\sum_{k=2}^{\infty}\frac1{(k-2)!}+4\sum_{k=2}^{\infty}\frac1{(k-1)!}$

$=\displaystyle\sum_{m=0}^{\infty}\frac1{m!}+4\left(\sum_{n=0}^{\infty}\frac1{n!}-\frac1{0!}\right)$

Now we know $\displaystyle\sum_{r=0}^{\infty}\frac{x^r}{r!}=e^x$

More generally if the numerator is of the form $\displaystyle a_0+a_1n+a_2n^2+a_3n^3+\cdots,$

we can express it as $\displaystyle b_0+b_1n+b_2n(n−1)+b_3n(n−1)(n−2)+\cdots$

-
– lab bhattacharjee Mar 30 '14 at 13:03
Thank you, very much finally got an answer 5e-4 (wolframalpha checked). – user139016 Mar 30 '14 at 15:57
@user139016, my pleasure. Sorry for the typo – lab bhattacharjee Mar 31 '14 at 14:25

We have, $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

$$xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+\cdots$$

$$(x+1)e^x = 1+2x+\frac{3x^2}{2!}+\cdots$$

At $x=1$ this gives you the first term $\sum\frac{k^2}{k!}$. The second term is $3(e-1)$(try to prove yourself).

-

First of all, the second part is quite trivial:

$$3\sum_{k=2}^\infty \frac{k}{k!}=3\sum_{k=2}^\infty \frac{1}{(k-1)!}$$ which you recognize as the sum that yields $e$, but without the zeroth term, so $3(e-1)$.

The first term is similar, you just have to split it into $k(k-1)+\text{something}$ to cancel out two terms of the factorial.

More generally, you could use generating functions. Introduce this as a power series at $x=1$. Then you can use integration and differentiation.

Edit: if you're supposed to use differentiation, this approach with series is the right one:

$$f(x)=\sum_{k=2}^\infty \frac{k^2+3k}{k!}x^{k-1}$$ Integrate once: $$\int_0^x f(y)dy=\sum_{k=2}^\infty \frac{k+3}{k!}x^k$$ Multiply by $x^2$ (to get the next integral to cancel out $k+3$: $$x^2 \int_0^x f(y)dy=\sum_{k=2}^\infty \frac{k+3}{k!}x^{k+2}$$ Integrate again: $$\int_0^x y^2 \int_0^y f(z)dz dy=\sum_{k=2}^\infty \frac{1}{k!}x^{k+3}$$

This you can evaluate: $$\sum_{k=2}^\infty \frac{1}{k!}x^{k+3}=x^3 \sum_{k=2}^\infty \frac{x^k}{k!}=x^3 (e^x-x-1)$$

Now differentiate, divide by $x^2$, differentiate again, and you get the original sum. Now put in $x=1$ to get the value you are looking for.

Note that this is a very long and unnecessary process because I pretended to be stupid to illustrate the point. In reality, you can recognize and regroup the terms: $k^2+3k=(k+1)k+2(k+1)-2$, split, and all three terms become trivial!

-