# Interchanging integration and summation vs. convergence

• September 21st 2010, 10:56 AM
Heirot
Interchanging integration and summation vs. convergence
I'm trying to calculate the series (all numerical data is calculated in Mathematica)

$S=\sum_{n=0}^{\infty}{\frac{1}{(n+1)(2n+1)(3n+1)}} \approx 1.05964$

I proceeded as follows. First the partial fraction decomposition of the summand,

${\frac{1}{(n+1)(2n+1)(3n+1)}=\frac{1}{2}\frac{1}{n +1}-4\frac{1}{2n+1}+\frac{9}{2}\frac{1}{3n+1}$,

then introducing the variable x:

$\frac{1}{(n+1)(2n+1)(3n+1)}=\frac{1}{2}\frac{x^{n+ 1}}{n+1}-4\frac{x^{2n+1}}{2n+1}+\frac{9}{2}\frac{x^{3n+1}}{ 3n+1}|_0^1$.

This is then recognized as the integral

$S=\sum_{n=0}^{\infty}{\int_0^1{(\frac{1}{2}x^n-4 x^{2n}+\frac{9}{2}x^{3n})dx}}$

This is still numerically equal to the first expression. Now, I want to interchange the summation and integration and sum the geometric series,

$S=\int_0^1{(\frac{1}{2}\frac{1}{1-x}-4 \frac{1}{1-x^2}+\frac{9}{2}\frac{1}{1-x^3})dx}$,

but this integral is numerically equal to $0.7980$.

Why aren't I allowed to sum the geometric series?
• September 21st 2010, 12:42 PM
Opalg
Quote:

Originally Posted by Heirot
I'm trying to calculate the series (all numerical data is calculated in Mathematica)

$S=\sum_{n=0}^{\infty}{\frac{1}{(n+1)(2n+1)(3n+1)}} \approx 1.05964$

I proceeded as follows. First the partial fraction decomposition of the summand,

${\frac{1}{(n+1)(2n+1)(3n+1)}=\frac{1}{2}\frac{1}{n +1}-4\frac{1}{2n+1}+\frac{9}{2}\frac{1}{3n+1}$,

then introducing the variable x:

$\frac{1}{(n+1)(2n+1)(3n+1)}=\frac{1}{2}\frac{x^{n+ 1}}{n+1}-4\frac{x^{2n+1}}{2n+1}+\frac{9}{2}\frac{x^{3n+1}}{ 3n+1}|_0^1$.

This is then recognized as the integral

$S=\sum_{n=0}^{\infty}{\int_0^1{(\frac{1}{2}x^n-4 x^{2n}+\frac{9}{2}x^{3n})dx}}$

This is still numerically equal to the first expression. Now, I want to interchange the summation and integration and sum the geometric series,

$S=\int_0^1{(\frac{1}{2}\frac{1}{1-x}-4 \frac{1}{1-x^2}+\frac{9}{2}\frac{1}{1-x^3})dx}$,

but this integral is numerically equal to $0.7980$.

Why aren't I allowed to sum the geometric series?

The thing that goes wrong is that you are not allowed to interchange the two limiting processes (integration, and summation of infinite series). This is only permitted under some condition like uniform convergence of the series. In this case, all three of the sums of the series ( $\frac{1}{1-x},\; \frac{1}{1-x^2},\;\frac{1}{1-x^3}$) give rise to divergent improper integrals, because they go to infinity at x=1. Interchanging summation and integration is definitely not safe in those conditions.

In fact, this calculation is a very nice example of why you need to be careful about interchanging limiting operations.
• September 21st 2010, 01:31 PM
emakarov
I agree. A sufficient condition for the interchange is the uniform convergence of the series on [0, 1] (see the last paragraph of the Integrability section of the Wikipedia article on Riemann integrals and this Google result in PDF). However, $\sum_{i=0}^n x^i=(1-x^{n+1})/(1-x)$, and though it converges to $1/(1-x)$ pointwise, it does not converge to it uniformly. Indeed, $1/(1-x) - (1-x^{n+1})/(1-x) = x^{n+1}/(1-x)\to_{x\to1}\infty$.
• September 21st 2010, 03:21 PM
Heirot
If I were to integrate to some epsilon, where the convergence is uniform, and then let epsilon -> 1, would that be allowed?
• September 22nd 2010, 08:33 AM
Heirot
It appears that even taking sum up to N, and at the very end taking N to be infinite doesn't work.