Radius and Interval of Convergence for (3x-2)^2/n 3^n Series

aydin

Homework Statement



find the radius of convergence and interval of convergence of the series


Σ (3x-2)^2 / n 3^n
n=1


Homework Equations





The Attempt at a Solution

 
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You need to clean up your notation a little bit, but the ROC for this series is easily found using any number of tests. Try the ratio test or the limsup test.
 
\sum_{n=1}^\infty \frac{(3x-2)^n}{n n!}
is, I think, what you want.

Ratio test is probably best.
a_n= \frac{|3x-2|^n}{n n!}
and
a_{n+1}= \frac{|3x-2|^{n+1}}{(n+1) (n+1)!}

so the ratio is
\frac{|3x-2|^{n+1}}{(n+1)(n+1)!}\frac{n n!}{|3x-2|^n}= |3x-2|\frac{n}{(n+1)^2}
What is the limit of that as n goes to infinity? If that is less than 1, the series will converge.
 
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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