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May 1st, 2018, 06:13 AM   #1
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interval / radius of convergence

Any guidance will be appreciated :

find the interval of convergence

u_n=((2^n (n+2))/n^2 )((2x-1)^n)
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May 1st, 2018, 10:20 AM   #2
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A pretty standard method for power series is to use the "ratio test".

Here $\displaystyle u_n= \frac{2^n(n+2)}{n^2}(2x- 1)^n$ and $\displaystyle u_{n+1}= \frac{2^{n+1}((n+1)+ 2)}{(n+1)^2}(2x- 1)^{n+1}$$\displaystyle = \frac{2(2^n)(n+ 3)}{(n+1)^2}(2x- 1)^{n+1}$

So $\displaystyle \frac{u_{n+1}}{u_n}=$$\displaystyle \frac{2^{n+1}}{2^n}\frac{n+3}{n+1}\frac{(n+1)^2}{n ^2}\frac{(2x- 1)^{n+1}}{(2x- 1)^n}$$\displaystyle = \frac{(2^{n+1}}{2^n}\frac{n+3}{n+1}\frac{(n+1)^2}{ n^2}(2x- 1)^1$

Take the limit as n goes to infinity, $\displaystyle \frac{2^{n+1}}{2^n}= 2$, $\displaystyle \frac{n+2}{n+1}$ goes to 1, $\displaystyle \frac{n+1)^2}{n^2}= \frac{n^2+ 2n+ 1}{n^2}= 1+ \frac{2}{n}+ \frac{1}{n^2}$ goes t0 1 so this all goes to $\displaystyle 2(2x- 1)$.

The series converges, absolutely, for the absolute value of that fraction less than 1: $\displaystyle |4x- 2|< 1$ so -1< 4x- 2< 1, 1< 4x< 3, 1/4< x< 3/4.
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May 1st, 2018, 09:04 PM   #3
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Thanks for the help!
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