Convergence of Series: Checking Equality by Changing Limits

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SUMMARY

The discussion centers on the equality of two series derived from the integral \(\int_{0}^{\infty}\frac{x^3}{e^x-1}dx\) which equals \(\frac{\pi^4}{15}\). The participant initially miscalculated the series representation, arriving at \(\sum_{n=-1}^{\infty}\frac{6}{n^4}\), incorrectly including a term that leads to division by zero. Upon reviewing their professor's work, they identified a substitution error that should have started the series at \(n=1\), confirming the series' equality with \(\sum_{k=1}^{\infty}\frac{1}{k^4} = \frac{\pi^4}{90}\).

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bob1182006
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Are these two series equal? (Solved)

Not a homework problem but didn't think other forums were a better place to post this.

I'm trying to show:
[tex]\int_{0}^{\infty}\frac{x^3}{e^x-1}dx=\frac{\pi^4}{15}[/tex]

After solving the integral, and checking it a few times, I get to this series:

[tex]\sum_{n=-1}^{\infty}\frac{6}{n^4}[/tex]

I don't know if I can just say that:

[tex]\sum_{n=-1}^{\infty}\frac{1}{n^4}=\sum_{k=1}^{\infty}\frac{1}{k^4}[/tex]

Since the RHS is equal to [itex]\pi^4 / 90[/itex]

If I try to just pull out the first 2 terms on the LHS then I get the sum of 1+1/0, which is bad..
 
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You definitely can't say those series are equal. There is no mathematically sound way of calculation you could have gotten the LHS standing on its own as part of a solution, since it includes 1/0. There must be an error in your calculations
 
Wow, thanks, just looked over what my professor did to start me off on this and noticed he made a mistake in a substitution and got n=-1 as the start of the series when it should have been n=1 so the series are equal!

Guess I should always look over everything, not just the part that I solved.
 

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