Solve Infinite Series: Sum of -(5/4)^n

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[SOLVED] Infinite series help

Homework Statement



\sum- (\frac{5}{4})^n
i=infinity and n=0

Homework Equations


Convergence of a geometric series
\sum (ar)^n = a/(1-r) when 0<|r|<1

The Attempt at a Solution


I have to explain why this series diverges or converges. The test for divergence gives an answer of infinity so it diverges. The terms are 1, -5/4, 25/16, -125/64, 625/256... To me it looks like a geometric series with r=|-5/4| which diverges because |-5/4|\geq 1. Is this correct?
 
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Correct.
 
thanks!
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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