$\sum_{n=1}^{\infty}\frac{-1}{n(-4)^n}=?$

Question

The problem asks to find the sum of $$ \sum\limits_{n=1}^{\infty} \frac{-1}{n(-4)^n}$$ and I tried to use the partial sums to find the answer and failed. Here is my take: $$ s_n=\frac{1}{1\times4^1}-\frac{1}{2\times4^2}+\frac{1}{3\times4^3}-\frac{1}{4\times4^4}+\cdots\\ \frac{1}{4}s_n= \frac{1}{1\times4^2}-\frac{1}{2\times4^3}+\frac{1}{3\times4^4}-\frac{1}{4\times4^5}+\cdots\\ s_n+\frac{1}{4}s_n=\frac{1}{1\times4^1}+(1-\frac{1}{2})\times\frac{1}{4^2}+\cdots$$ and I am completely lost.

Answer 1

The standard trick here is to use the geometric series (since you in some sense just have $-1/4$ raised to the power $n$ in each term) and manipulate it with derivatives, or, in this case, integrals. Observe that

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

Start at $n=1$ instead:

$$\sum_{n=1}^\infty x^{n-1} = \frac{1}{1-x}$$

Integrate termwise:

$$C+\sum_{n=1}^\infty \frac 1 n x^n = - \ln(1-x)$$

Observe that $x=0$ justifies choosing $C=0$:

$$\sum_{n=1}^\infty \frac 1 n x^n = - \ln(1-x)$$

Now observe what happens when you plug in $x=-1/4$ and compare to your own situation.

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Generally, a trick like this can be applied to many sums of the type

$$\sum_{n=0}^\infty n^k x^n$$

for $k$ an integer: it will require sufficient use of derivatives ($k>0$) or integrals ($k<0$) and some trivial algebraic manipulations to make it work though. The underlying key that makes this easy is that the geometric series is absolutely convergent (so differentiation/integration may be done termwise).