# Sum calculation

1. Mar 17, 2008

### sukreth

Im not sure if it is related to calculus but,

Calculate the sum

$$\sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!}$$

exactly.

I tried to to partial fraction decomposition but couldn't find anything.

Last edited: Mar 17, 2008
2. Mar 17, 2008

### lurflurf

this problem is the x=1 case of
$$\sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!}x^n=\sum^{\infty}_{n=0}[(xD)^2-1]\frac{x^n}{n!}=[(xD)^2-1]\sum^{\infty}_{n=0}\frac{x^n}{n!}=[(xD)^2-1]e^x=(x^2+x-1)e^x$$
in particular for x=1
(x^2+x-1)e^x|x=1=e
uniform convergence was used to bring the derivative out of the infinite sum

Last edited: Mar 17, 2008
3. Mar 18, 2008

### sukreth

I think this solution is beyond my level of knowledge. How did you get (n-1)(n+1) out of the sum? I sadly don't know about uniform convergence. Is there any way to do this with simpler algebraic methods?

4. Mar 18, 2008

### Mute

Simply expand the numerator to start - it's a difference of squares, n^2 -1; then, split the expression up into two sums:

$$\sum_{n=0}^{\infty} \frac{n^2-1}{n!} = \sum_{n=0}^{\infty}\frac{n^2}{n!} - \sum_{n=0}^{\infty}\frac{1}{n!}$$

You should recognize the last sum as the Taylor series for e^x at x=1. In the first, you cancel a factor of n:

$$\sum_{n=1}^{\infty}\frac{n}{(n-1)!}$$

Now, note that the first term of this sum starts at 1, since n^2/n! is zero when n = 0, so we write n/(n-1)! only after we have accounted for that fact. Now, to evaluate this sum, relabel the index of your sum by a new variable, m for instance, and let m = n - 1. Rewrite the sum in terms of m. The remaining steps to arrive at the answer are similar to those already performed, so you should hopefully be able to find the solution from here.

5. Mar 19, 2008

### rwinston

Infinite Sums Product

is it possible to do the following:

$$\sum_{m=0}^\infty{\frac{n}{m!}} \equiv \sum_{n=0}^\infty{n} \sum_{m=0}^\infty{\frac{1}{m!}}?$$

Last edited: Mar 19, 2008
6. Mar 19, 2008

### tiny-tim

Welcome to PF!

Hi sukreth! Welcome to PF!

A little trick, based on what Mute said, but made a little neater for you to follow:

You obvously understand partial fractions, so imagine you ignore all the terms in the denominator after the first two.

That gives you (n+1)(n-1)/n(n-1),
which you know how to turn into:
{n(n-1) + n + (-1)}/n(n-1),
= 1 + 1/(n-1) - 1/n(n-1).​

Of course, this is only valid for n ≥ 2.

So, putting back in the rest of the denominator, which is (n-2)!, and treating the n= 0 and 1 cases separately, and remembering that 0! = 1, makes the original
$$\sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!}$$​
equal to
$$-1\,+\,0\,+\,\sum^{\infty}_{n=2}\frac{(n-1)(n+1)}{n!}$$
$$=\,-1\,+\,\sum^{\infty}_{n=2}\frac{1}{(n-2)!}\,+\,\sum^{\infty}_{n=2}\frac{1}{(n-1)!}\,-\,\sum^{\infty}_{n=2}\frac{1}{n!}$$
$$=\,-1\,+\,\sum^{\infty}_{n=0}\frac{1}{n!}\,+\,\sum^{\infty}_{n=1}\frac{1}{n!}\,-\,\sum^{\infty}_{n=2}\frac{1}{n!}$$
= -1 +e + (e - 1) - (e - 2) = e.​

In fact, using the same process, you can prove:
$$\sum^{\infty}_{n=0}\frac{(n-1)(n+1)x^n}{n!}$$
$$=\,-1 +e^x + (e^x\,-\,1) - (e^x\,-\,1\,-\,x) = e^x\,+x\,-1\,.$$​

All this is without any calculus, and uses only the definition of e (or e^x).

7. Mar 19, 2008

### rwinston

Nice!

just as a matter of interest, is it possible in principle to separate an infinite sum like I have shown above? Or is this incorrect?

8. Mar 19, 2008

### Mute

What you wrote above is incorrect unless you forgot to include a sum over all n on the left hand side of the equation. What is true is that

$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(n)g(m) = \left(\sum_{n=0}^{\infty}f(n)\right)\left(\sum_{m=0}^{\infty}g(m)\right)$$

Note that whatever your summand (the thing in the sum) is, it must be factorizable into two separate functions, on of n and one of m, in order for you to treat the double sum as a product of sums, AND the upper limit of one sum cannot depend on the other variable. That is,

$$\sum_{n=0}^{\infty}\sum_{m=0}^{h(n)}f(n)g(m) \neq \left(\sum_{n=0}^{\infty}f(n)\right)\left(\sum_{m=0}^{h(n)}g(m)\right)$$

where the upper limit of the sum over m is some function of n. In essence, when the two sums are entirely independent of one another you can treat the double sum as a product of sums. When one sum depends on the other sum's variable somehow, they cannot be treated as such.

9. Mar 19, 2008

### lurflurf

This is a simple algebraic method.
The ideal is consider a sum
$$S[p(n)]:=\sum^{\infty}_{n=0}\frac{x^n}{n!}p(n)$$
where p(n) is a polynomial as in your example
since a polynomial is a linear combination of powers
ie 3n^2+4n+7 or n^2-1 in your example
we can consider
$$S[n^k]:=\sum^{\infty}_{n=0}\frac{x^n}{n!}n^k$$
now consider an operatior
xD:R[x]->R[x]
ie take the derivative with respect to x and multiply by x
xD has a simple and interesting effect on x^n
xDx^n=n*x^n
it can also be repeated
[(xD)^k]x^n=(n^k)*(x^n)
applied to polynomials
p(xD)x^n=p(n)*x^n
so this operator is just what we want
a note
we must be sure to respect the product rule
(xD)^0=1
(xD)^1=xD
(xD)^2=(x^2)(D^2)+xD
(xD)^3=(x^3)(D^3)+3(x^2)(D^2)+xD
(xD)^4=(x^4)(D^4)+6(x^3)(D^3)+7(x^2)(D^2)+xD
and so on
now we can solve the problem
$$S[p(n)]=\sum^{\infty}_{n=0}p(n)\frac{x^n}{n!}=\sum^{\infty}_{n=0}p(xD)\frac{x^n}{n!}=p(xD)\sum^{\infty}_{n=0}\frac{x^n}{n!}=(p(xD))S[1]=p(xD)e^x$$
The only possible problem with this procedure is that we have brought an opperator out of an infinite sum (commuted them)
This can be a problem if the sum does not have "good" convergence
but we know the series for e^x converges uniformly so all is well
we could also take this on faith
or we could prove S[p(n)]=g(x)e^x
for some polynomial g of the same order as p
and our method produces it
so the given problem can be written
S[n^2-1]|x=1
[(xD)^2-1]e^x|x=1
[(x^2)D^2+xD-1]e^x|x=1
(x^2+x-1)e^x|x=1
(1+1-1)e^1
1*e
e