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Convergence of an Infinite series and a related Qn

bincybn

Member
Apr 29, 2012
36
Dear friends,

\(\displaystyle \sum_{x=1}^{\infty}\frac{1}{x}\) diverges.

But \(\displaystyle \sum_{x=1}^{\infty}\frac{1}{x^{2}}=\frac{\pi^{2}}{6}\)

How can we prove that \(\displaystyle \sum_{x=1}^{\infty}\left(\frac{1}{x^{\left(1+epsilon\right)}}\right)\) converges to a finite value?



Thanks in advance.

Bincy.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I preface every post like this that I could very well be wrong, but here is my take on it before someone else can confirm/deny my reasoning or provide a different proof.

If epsilon is an integer then I think induction can prove this.

1) Looking at \(\displaystyle \frac{1}{x^n}\) you know it converges for n=2. I suppose your question is looking at \(\displaystyle \frac{1}{x^{n+1}}\), so n=1 is true.

2) Using the ratio test, you can show that if n converges that implies that (n+1) converges.
 
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bincybn

Member
Apr 29, 2012
36
I agree whatever you said. But there is a small catch.By epsilon, I meant that a very small real no. like 10^-10. For n>=2, we can prove the convergence of the series. But what about 1<n<2 ? If we can prove the convergence for n=1+ (1+ means epsilon greater than 1), any infinite series of this kind converges for n>1.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I think you can use the integral test for convergence.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041

dwsmith

Well-known member
Feb 1, 2012
1,673
I believe there is a nice proof in Lang's Complex Analysis of the Riemann Zeta Function. If not, I have a proof from Foote.