# Closed form of $\sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$

Let $$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$$ 1. Find the radius of convergence.
2. Calculate $S(x)$.
3. Find $S^{(n)}(x)$ without computing the derivatives of $S(x)$.

From the root test I find $R = 1/4$. It's the second point that troubles me. This is my attempt: \begin{align} S(x) &= \sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n =\\ &= \sum\limits_{n=1}^\infty 2^{2n+1}\int_{-4}^x (t + 4)^{2n-1}\mathrm dt =\\ &= \int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n+1}(t + 4)^{2n-1}\mathrm dt =\\ &= 4\int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n-1}(t + 4)^{2n-1}\mathrm dt =\\ &=\ ??? \end{align}

I don't know how to continue from there. I know that I should transform the inner sum into a known Taylor expansion or a geometric series, but I don't see how I could do that.

As for the last point, we have that $$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n = \sum_{n = 0}^\infty \frac{S^{(n)}(x)}{n!}(x + 4)^n,$$ as per the Taylor series definition. However, I don't know how to reconcile the indices and the two powers $2n$ and $n$.

• One thing to note is that $4^n (x+4)^{2n} = (4(x+4)^2)^n$. – Mark B Feb 9 '16 at 20:49
• And for part 2, finding $S'(x)$ will make the above much more useful. – DanielV Feb 9 '16 at 20:50

Let's set $A = 4(x+4)^2$. You want to find $$\sum\limits_{n=1}^{\infty} \frac{1}{n}A^n = \sum\limits_{n=1}^{\infty} \int_{0}^A t^{n-1}\,dt = \int_{0}^A \sum\limits_{n=1}^{\infty} t^{n-1} \,dt = \int_0^A \frac{1}{1-t}\,dt = \cdots$$

• How do you justify interchanging the series with the integral? – Mark Viola Feb 9 '16 at 21:59
• Inside the radius of convergence the term-wise derivative and anti-derivative are the same as for the series. – LutzL Feb 9 '16 at 22:46
• The convergence interval of the new series is $[-1, 1)$. Then both $0$ and $A$ are inside the convergence interval and the swap between the series and the integral is justified. Am I correct? – rubik Feb 10 '16 at 8:58
• Since $A \geq 0$, then the integrand $t^{n-1}$ only takes on non-negative values. So, you can interchange the sum and integral using Tonelli's theorem. – Tom Feb 11 '16 at 14:44

A quick trick to compute $S(x)$ is the following. Use $$\frac{1}{n}=\int_0^\infty ds\ e^{-sn}$$ to write $$S(x)=\int_0^\infty ds \sum_{n=1}^\infty (4e^{-s} (x+4)^2)^n=\int_0^\infty ds \frac{4e^{-s} (x+4)^2}{1-4e^{-s} (x+4)^2}=-\ln \left(-4 x^2-32 x-63\right)\ ,$$ where one uses the geometric series and the simple substitution $e^{-s}=z$.

Let $t=4(x+4)^2$ and the series becomes

$$\sum_{k=1}^\infty\frac{t^n}n.$$

You can recognize the Taylor development of $-\ln(1-t)$, or derive the series to get

$$\sum_{k=1}^\infty t^{n-1}=\frac1{1-t}.$$

I'd try the following:

$$|x|<1\implies \frac1{1-x}=\sum_{n=0}^\infty x^n\stackrel{\text{diff.}}\implies\frac1{(1-x)^2}=\sum_{n=1}nx^{n-1}$$

Check now that $\;|4(x+4)^2|<1\;$ for $\;|x+4|<\frac14\;$ and you get $\;S(x)\;$ after integrating (or integrating directly)

The approach in the OP can work. We need only make a simple substitution to facilitate. To that end, we proceed.

Let $S_N(x)$ denote the partial sums of the series $S(x)=\sum_{n=1}^\infty \frac{4^n(x+2)^{2n}}{n}$. Then, letting $y=(2x+8)^2$, with $|y|<1$, we have

\begin{align} S_N(x)&=\sum_{n=1}^N \frac{4^n(x+2)^{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 -\log|1-y|\,\,\text{as}\,\,N\to \infty \tag 1\\\\ &=-\log\left|1-(2x+8)^2\right|\\\\ &=-\log|2x+9|-\log|2x+7| \end{align}

where the justification for interchanging the limit with the integral in $(1)$ is provided by the Dominated Convergence Theorems since $\left|\frac{1-z^N}{1-z}\right|\le \frac{2}{|1-z|}$ for $|z|<1$.

Let's put $\alpha:=4(x+4)^2$; we want to find $$\sum_{n=1}^{+\infty}\frac{\alpha^n}{n}$$

We want to compute the following sum: $$\sum_{n=1}^{+\infty}\frac{1}{nz^n},\;\;\;\; z\in\mathbb C\;.$$ We immediately see that $|z|>1$, in order to have absolute convergence.

We recall first two results:

$\bullet\;\;$First: $$\log(1+z)=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{z^n}{n},\;\;\;\forall |z|<1$$ $\bullet\;\;$Second: $$\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)= \sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1$$ The last one can be proved, showing by induction that $\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}$.

Ok: \begin{align*} \sum_{n=1}^{+\infty}\frac{1}{nz^n}=& \sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{1}{z}\right)^n\\ =&\underbrace{\sum_{k=0}^{+\infty}\frac{1}{2k+1}\left(\frac{1}{z}\right)^{2k+1}- \sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}}_{\log\left(1+\frac{1}{z}\right)}+ 2\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}\\ =&\log\left(1+\frac{1}{z}\right)+ \sum_{k=1}^{+\infty}\frac{1}{k}\left(\frac{1}{z^2}\right)^{k}\\ =&\log\left(1+\frac{1}{z}\right)+ \log\left(1+\frac{1}{z^2}\right)+\cdots\\ =&\sum_{n=0}^{+\infty}\log\left(1+\frac{1}{z^{2^n}}\right)\\ =&\log\left(\prod_{n=0}^{+\infty}\left(1+\left(\frac{1}{z}\right)^{2^n}\right)\right)\\ =&\log\left(\frac{z}{z-1}\right) \end{align*}

just put now $z=1/\alpha$ and conclude.