# Efficient method to evaluate the following series

How do I calculate the infinite series:

$$\frac{1^2\cdot 2^2}{1!}+\frac{2^2\cdot 3^2}{2!}+\dots \quad?$$ I tried to find the nth term $t_n$.$$t_n=\frac{n^2\cdot (n+1)^2}{n!}.$$ So, $$\sum_{n=1}^{\infty}t_n=\sum_{n=1}^{\infty}\frac{n^4}{n!}+2\sum_{n=1}^{\infty}\frac{n^3}{n!}+\sum_{n=1}^{\infty}\frac{n^2}{n!}$$ after expanding. But I do not know what to do next.

Thanks.

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One approach, not necessarily best, is to cancel n, then split remaining n as (n-1)+1 and cancel n-1, and keep going (paying attention to the beginning of the series). You get some integers and some e. – user31373 Jul 18 '12 at 3:19
We can write $$n^2(n+1)^2 = an + bn(n-1) + cn(n-1)(n-2) + dn(n-1)(n-2)(n-3).$$ Comparing coefficients yields $$a = 4, \quad b = 14, \quad c = 8, \quad d = 1.$$ Then the rest is clear, yielding $$\sum_{n=1}^{\infty} \frac{n^2(n+1)^2}{n!} = (a+b+c+d)e = 27e.$$ – Sangchul Lee Jul 18 '12 at 3:20
Look up Dobiński's Formula. – J. M. Jul 18 '12 at 3:32
@sos440 I like your method much. – Lingqiao Jul 18 '12 at 3:41

You are right, now you need to expand them separately and express each of them in form of $e$:

$$\sum \limits_{n=1}^{\infty}\frac{n^2}{n!}= \sum \limits_{n=1}^{\infty} \frac{n+(n-1)n}{n!} = \sum \limits_{n=1}^{\infty} \frac 1{(n-1)!} +\frac 1{(n-2)!} = 2e$$

Similarly, we can show that,

$$\sum \limits_{n=1}^{\infty}\frac{n^3}{n!}= \sum \limits_{n=1}^{\infty} \frac{n+(n^2-1)n}{n!} = 5e$$

and,

$$\sum \limits_{n=1}^{\infty}\frac{n^4}{n!}= 15e$$

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thanks.I like this method. – Lingqiao Jul 18 '12 at 3:40

The forms of the power series, you reached in the last step of your work, are known as Bell numbers. Bell numbers are defined as:

$$B_{m} = \frac{1}{e}\,\sum_{n=0}^{\infty} \frac{n^m}{n!}$$.

So your result can be written in terms of Bell numbers:

$$e( B_{4} + 2\,B_{3} + B_{2})\,,$$

where $B_{4}=15 \,, B_{3}=5\,, B_{2}=2\,.$ See here.

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$$t_n = \frac{4}{n!} \left( \frac{n(n+1)} {2} \right)^2$$

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tn = (4/n!)[n(n+1)/2]^2 – Rajesh K Singh Jul 18 '12 at 8:29

Given that $P(x)$ is a polynomial of degree $n$, we have $$P(x)=\sum_{k=0}^n\binom xk\Delta^kP(0)$$ which is Newton's series. Therefore \begin{align} \sum_{m\ge0}\frac{P(m)}{m!}&=\sum_{m\ge0}\sum_{k=0}^n\frac{\Delta^kP(0)}{k!(m-k)!}\\ &=\sum_{k=0}^n\frac{\Delta^kP(0)}{k!}\sum_{m\ge k}\frac1{(m-k)!}\\ &=e\sum_{k=0}^n\frac{\Delta^kP(0)}{k!} \end{align}

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