Yes, Use P-Series Test to Confirm Series Convergence

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if my an = 1 / (n+1)^2
can i use p series test to confirm that the series will converge..
where p = 2..
 
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Sure. If your series is something like this,
\sum_{n = 0}^{\infty} \frac{1}{(n + 1)^2}

you can change the series index to write the series this way:
\sum_{n = 1}^{\infty} \frac{1}{n^2}
 
it's comparison test right?
the series will converge..
am i right?
 
p-series converge for p>1, so yes, your summation converges.
 
naspek said:
it's comparison test right?
the series will converge..
am i right?
You don't need the comparison test if you know the p-series test, and this is a p-series.
 
I believe that naspek was saying that he can argue that 1/(n+1)^2< 1/n^2 and so use the comparison test without using your idea of changing the index.
 
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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