MHB Convergence of an Infinite series and a related Qn

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The discussion centers on the convergence of the series \(\sum_{x=1}^{\infty}\left(\frac{1}{x^{(1+\epsilon)}}\right)\) for small positive values of epsilon. It is established that the series diverges for \(\epsilon = 0\) and converges for \(\epsilon \geq 1\) using induction and the ratio test. The integral test is suggested as a method to prove convergence for values of \(n\) between 1 and 2. References to resources like Wikipedia and Lang's Complex Analysis are made for further proof and understanding. The conversation emphasizes the importance of epsilon in determining the convergence of the series.
bincy
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Dear friends,

[math] \sum_{x=1}^{\infty}\frac{1}{x}[/math] diverges.

But [math] \sum_{x=1}^{\infty}\frac{1}{x^{2}}=\frac{\pi^{2}}{6}[/math]

How can we prove that [math] \sum_{x=1}^{\infty}\left(\frac{1}{x^{\left(1+epsilon\right)}}\right)[/math] converges to a finite value?
Thanks in advance.

Bincy.
 
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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 [math]\frac{1}{x^n}[/math] you know it converges for n=2. I suppose your question is looking at [math]\frac{1}{x^{n+1}}[/math], 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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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.
 
I think you can use the integral test for convergence.
 
Evgeny.Makarov said:
I think you can use the integral test for convergence.

Interestingly Wikipedia uses the OP's problem as an example of the integral test.
 
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.
 
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