Summation: Calculating an Infinite Series

In summary, the infinite sum \sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!} can be solved using partial fraction decomposition and the concept of uniform convergence. This involves expanding the numerator into a difference of squares, splitting the expression into two sums and using the Taylor series for e^x. The resulting solution is -1 +e + (e - 1) - (e - 2) = e. Alternatively, the problem can be solved using an operator xD which has the effect of multiplying a polynomial by n, and then using the series for e^x which has been proven to converge uniformly.
  • #1
sukreth
3
0
Im not sure if it is related to calculus but,

Calculate the sum

[tex]\sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!}[/tex]

exactly.

I tried to to partial fraction decomposition but couldn't find anything.
 
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  • #2
this problem is the x=1 case of
[tex]\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[/tex]
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
 
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  • #3
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
Simply expand the numerator to start - it's a difference of squares, n^2 -1; then, split the expression up into two sums:

[tex]\sum_{n=0}^{\infty} \frac{n^2-1}{n!} = \sum_{n=0}^{\infty}\frac{n^2}{n!} - \sum_{n=0}^{\infty}\frac{1}{n!}[/tex]

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:

[tex]\sum_{n=1}^{\infty}\frac{n}{(n-1)!}[/tex]

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
Infinite Sums Product

is it possible to do the following:

[tex]
\sum_{m=0}^\infty{\frac{n}{m!}} \equiv \sum_{n=0}^\infty{n} \sum_{m=0}^\infty{\frac{1}{m!}}?
[/tex]
 
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  • #6
Welcome to PF!

Hi sukreth! Welcome to PF! :smile:

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
[tex]\sum^{\infty}_{n=0}\frac{(n-1)(n+1)}{n!}[/tex]​
equal to
[tex]-1\,+\,0\,+\,\sum^{\infty}_{n=2}\frac{(n-1)(n+1)}{n!}[/tex]
[tex]=\,-1\,+\,\sum^{\infty}_{n=2}\frac{1}{(n-2)!}\,+\,\sum^{\infty}_{n=2}\frac{1}{(n-1)!}\,-\,\sum^{\infty}_{n=2}\frac{1}{n!}[/tex]
[tex]=\,-1\,+\,\sum^{\infty}_{n=0}\frac{1}{n!}\,+\,\sum^{\infty}_{n=1}\frac{1}{n!}\,-\,\sum^{\infty}_{n=2}\frac{1}{n!}[/tex]
= -1 +e + (e - 1) - (e - 2) = e.​

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

All this is without any calculus, and uses only the definition of e (or e^x). :smile:
 
  • #7
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
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

[tex]\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)[/tex]

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,

[tex]\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)[/tex]

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
sukreth said:
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?
This is a simple algebraic method.
The ideal is consider a sum
[tex]S[p(n)]:=\sum^{\infty}_{n=0}\frac{x^n}{n!}p(n)[/tex]
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
[tex]S[n^k]:=\sum^{\infty}_{n=0}\frac{x^n}{n!}n^k[/tex]
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
[tex]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[/tex]
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
 

1. What is a summation?

A summation is a mathematical operation that involves adding a sequence of numbers or terms together. It is represented by the symbol Σ (sigma) and is also known as a series or a finite sum.

2. What is an infinite series?

An infinite series is a summation where the number of terms being added is infinite. This means that the series continues indefinitely without an endpoint. It is often represented by the symbol ∞ (infinity).

3. How do you calculate an infinite series?

The calculation of an infinite series involves finding the sum of all the terms in the series. This can be done using various mathematical techniques such as the geometric series formula, the telescoping series method, or the use of convergence tests such as the ratio test or the integral test.

4. Why is calculating an infinite series important?

Calculating an infinite series is important in mathematics as it helps us understand the behavior of sequences and series. It also has applications in various fields such as physics, engineering, and economics, where infinite series can be used to model real-world situations and make predictions.

5. What is the difference between a convergent and a divergent series?

A convergent series is one in which the sum of all the terms approaches a finite value as the number of terms increases. In contrast, a divergent series is one in which the sum of the terms increases without bound as the number of terms increases. In other words, a convergent series has a finite sum, while a divergent series does not.

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