Solving a Puzzling Math Series: Need Help!

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Homework Statement



heres the problem: http://img527.imageshack.us/img527/9660/66702982oe9.png


Homework Equations



p series or geometric series?

The Attempt at a Solution



I thought this was a p series, but i was told it was a geometric series?

anyone who can walk me through this problem, please.
 
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This is a geometric series. A geometric series is in the form \sum_n x^n, while a P-Series is in the form, \sum_n n^p. Ignoring the factor of 3, you can solve this series using
\sum_{k=101}^\infty 5^{-k} = \sum_{k=0}^\infty 5^{-k} - \sum_{k=0}^{100} 5^{-k}. Both right hand terms should have analytic expressions I think.
 
In any geometric series all you need to compute the sum is the first term and the common ratio. The sum is always (first term)/(1-common ratio)
 
i know I'm close, i got 3/4 to be the sum.

heres a pic of what I've done, please correct me.

http://img518.imageshack.us/img518/3712/57667609ii9.png
 
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In your original question, the numerator was not being exponentiated to the k-th power. In your evaluation it is.

It may be best if you rewrite the series as 3 \sum_{k=101}^{\infty} \left( \frac{1}{5}\right)^k, and then take into account Mathdopes post.
 
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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