# Let S(x)=\sum_{n=1}^\infty\frac{4^n(x+4)^{2n}}{n} Find the radius of convergence.

Let
$S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{4}^{n}{\left(x+4\right)}^{2n}}{n}$
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Cicolinif73
Let ${S}_{N}\left(x\right)$ denote the partial sums of the series $S\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{4}^{n}{\left(x+2\right)}^{2n}}{n}$. Then, letting $y={\left(2x+8\right)}^{2}$, with $|y|<1$, we have
${S}_{N}\left(x\right)=\sum _{n=1}^{N}\frac{{4}^{n}{\left(x+2\right)}^{2n}}{n}$
$=\sum _{n=1}^{N}\frac{{y}^{n}}{n}$
$=\sum _{n=1}^{N}{\int }_{0}^{y}{z}^{n-1}dz$
$={\int }_{0}^{y}\frac{1-{z}^{N}}{1-z}dz$
$\to -\mathrm{log}|1-y|$ as $N\to \mathrm{\infty }$ (1)
$=-\mathrm{log}|1-{\left(2x+8\right)}^{2}|$
$=-\mathrm{log}|2x+9|-\mathrm{log}|2x+7|$
where the justification for interchanging the limit with the integral in (1) is provided by the Dominated Convergence Theorems since $|\frac{1-{z}^{N}}{1-z}|\le \mid \frac{2}{|1-z|}$ for $|z|<1$