1. When is it ok to interchange integration with summation to infinity?

2. Originally Posted by Heirot
When is it ok to interchange integration with summation to infinity?
If the partial sums of the summation converge uniformly.

In your case, a way to test for uniform convergence is the Weierstrass M test.

3. Thanks to all of you. I had a sum a while ago which to this day I cannot perform: 1/((n+1)(2n+1)(3n+1)), n=0,...,infinity. I tried using the trick of interchanging the summation with integration there and it didn't work. I'm wondering, is there another way of doing this sum?

4. Originally Posted by Heirot
Thanks to all of you. I had a sum a while ago which to this day I cannot perform: 1/((n+1)(2n+1)(3n+1)), n=0,...,infinity. I tried using the trick of interchanging the summation with integration there and it didn't work. I'm wondering, is there another way of doing this sum?
$\displaystyle \frac1{(n+1)(2n+1)(3n+1)} = \frac a{n+1}+\frac b{2n+1}+\frac c{3n+1}$

So $\displaystyle \sum_{n=0}^\infty \frac1{(n+1)(2n+1)(3n+1)} = \int_0^1 \sum_{n=0}^\infty \left(ax^n+bx^{2n}+cx^{3n}\right) dx = \int_0^1 \left(\frac{a}{1-x}+\frac{b}{1-x^2}+\frac{c}{1-x^3}\right) dx$ for $|x|<1$.

Can you take it from here?

5. I tried that - it doesn't work because the series don't converge uniformly so the integration and summation don't commute.

Mathematica gives two different numerical answers.

7. Originally Posted by Soroban
Hello, Heirot!

The general term is: . $a_n \;=\;\dfrac{1}{n(n+1)(n+2)(n+3)}$

Partial fractions: . $\displaystyle \frac{1}{n(n+1)(n+2)(n+3)} \;=\;\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} + \frac{D}{n+3}$

We find that: . $A = \frac{1}{6},\;B = -\frac{1}{2},\;C = \frac{1}{2},\;D = -\frac{1}{6}$

Hence: . $\displaystyle a_n \;=\;\frac{\frac{1}{6}}{n} - \frac{\frac{1}{2}}{n+1} + \frac{\frac{1}{2}}{n+2} - \frac{\frac{1}{6}}{n+3} \;=\;\frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

Then: . $\displaystyle S \;=\;\frac{1}{6}\sum^{\infty}_{n=1}\left(\frac{1}{ n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

Crank out several terms:

$\displaystyle S \;=\;\frac{1}{6}\bigg[\left(1 - \frac{3}{2} + \frac{3}{3} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{3}{3} + \frac{3}{4} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right)$

. . . . . . . $\displaystyle + \left(\frac{1}{4} - \frac{3}{5} + \frac{3}{6} - \frac{1}{7}\right) + \left(\frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8}\right) + \left(\frac{1}{6} - \frac{3}{7} + \frac{3}{8} - \frac{1}{9}\right)$

. . . . . . . . . $\displaystyle + \left(\frac{1}{7} - \frac{3}{8} + \frac{3}{9} - \frac{1}{10}\right) + \left(\frac{1}{8} - \frac{3}{9} + \frac{3}{10} - \frac{1}{11}\right) + \hdots \bigg]$

We find that nearly all the terms cancel out.

And we are left with: . $\displaystyle S \;=\;\frac{1}{6}\left(1 - \frac{3}{2} + \frac{1}{2} + \frac{1}{3}\right) \;=\;\frac{1}{18}$

I think I got you turned on to partial fractions (it's a wonderful method).

8. Hello,

I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

I'd like to express it as an integral, how should I proceed?

9. Originally Posted by Heirot
Hello,

I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

I'd like to express it as an integral, how should I proceed?
Well $\displaystyle\frac{1}{(1+a)\cdot (1+2a)...\cdot (1+k\cdot a)} = \frac{\left(\tfrac{1}{a}\right)^k}{\left(\tfrac{1} {a}+1\right)\cdot ... \cdot \left(\tfrac{1}{a}+k\right)}$

Now note that we have $\displaystyle\frac{1}{\left(\tfrac{1}{a}+1\right)\ cdot ... \cdot \left(\tfrac{1}{a}+k\right)} = \frac{\Gamma \left( \tfrac{1}{a} + 1\right)}{\Gamma(\tfrac{1}{a} + k + 1) } = \frac{\text{B}\left(\tfrac{1}{a}+1,k\right)}{\Gamm a( k)}$

Try to go on from there.

10. ## Comment and a question for you

Originally Posted by Heirot
Hello,

I have another summation problem: 1/(1+a) + 1/((1+a)(1+2a)) + 1/((1+a)(1+2a)(1+3a)) + ...

I'd like to express it as an integral, how should I proceed?
My comment is that normally the integral is given first from which a series is derived. Here you have it in reverse which makes the problem interesting.

My question is how do you know that the integral exists?

11. My approach to complicated sums is to express them as integrals. Being a physicist, I find it easier to think in terms of integrals than sums. You can always transform a number into a definite integral. The question is to choose the simplest integral. This is where I sometimes need help from you guys

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