What is the interval of convergence of $sum_1^oo (3x-2)^(n)/(1+n^(2)$ ?

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Answer 1 · 2016-08-22T16:51:30

Convergent for $1/3 < x < 1$

Call $S = sum_1^oo y^(n)/(1+n^(2))$ then

$sum_1^oo abs(y^n)/(1+n^(2)) < sum_1^oo abs(y^n)/n^2$ so

if $sum_1^oo abs(y^n)/n^2$ converges then $sum_1^oo abs(y^n)/(1+n^(2))$ also converges.

Calling $S_1 = sum_1^oo y^n/n^2$ we have

$y(d/(dy)S_1)=sum_1^oo y^n/n$ and

$y d/(dy)(y(d/(dy)S_1)) = sum_1^oo y^n$

supposing now that $abs(y) < 1$ we have

$y d/(dy)(y(d/(dy)S_1)) =1/(1-y)-1 = y/(1-y)$

and successively

$d/(dy)(y(d/(dy)S_1)) =1/(1-y)$
$y(d/(dy)S_1) =-log_e(1-y)$
$d/(dy)S_1 =-log_e(1-y)/y$
$S_1 = "PolyLog"(2,y)$

So for $abs (3x-2) < 1$ this series is convergent with convergence radius

$1/3 < x < 1$

Attached a plot showing $S$ in black and $S_1, "PolyLog(2,y)"$ in green and dashed red respectively.

enter image source here

What is the interval of convergence of sum_1^oo (3x-2)^(n)/(1+n^(2) sum_1^oo (3x-2)^(n)/(1+n^(2) ?