Finding Values of p for Convergence: (-1)^(n-1)/ n^p

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Homework Statement


For what values of p is the series con.
(-1)^ (n-1)/ (n)^(p)


Homework Equations



In order for a p series to converge, p should be greater than 1

The Attempt at a Solution



So an = 1/ n^p

and p > 1

Is that right ?
 
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You're making so many abbreviations and omissions that it's hard to tell what you're thinking -- I think you even confused yourself because of it. Try again, but this time say what you're doing, and why.
 
I don't think so,, the only abbreviation is Con. which means Convergent

** Am trying to find the valu of p for which the series will converge

What I thought about is : becauce it 's a p-series , then p should be P greater than 1 ?

What do you think ??
 
remaan said:
becauce it 's a p-series
No it's not.
 
So, give a hint? !
 
Oh, you I mean this part is a p- series, the one I considered as an
 
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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