Does the series of 1/n^(1+a) converge or diverge

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Homework Statement



I have to determine whether the infinite sum (from n=1) of 1/n^(1+a) converges or diverges. Where 0 < a < 1


Homework Equations



Ration Test

Comparison / Integral test


The Attempt at a Solution



I have tried using the ratio test but it gives me 1, so it cannot determine convergence or divergence.

I then tried using comparision test, comparing it to 1/n (which diverges) and 1/n^2 (which converges). But since 1/n^(1+a) is smaller than 1/n and greater than 1/n^2 I cannot make such comparison.

Then I used integral test where I let a=0.1 . This method tells me that for (a) greater than or equal to 0.1 the series converges. But how do I show that it converges (I think it converges) for (a) less than 0.1?

Thanks
 
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Why "a= 0.1"? Why not just leave it as "a"?
 


Hi thanks for the reply.

Leaving it as (a), I got the answer as

(1/(-2-a)).((1/n^(2+a)) - 1)

If this is right, then it will converge to 1.

So am does this mean that the series will converge to 1 as well?

thanks again
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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