# How to evaluate $\sum_{n=1}^\infty \frac{1}{n 2^{2n}}$?

Evaluate $$\sum_{n=1}^\infty \frac{1}{n2^{2n}}$$

I'd be glad for guidance, because, frankly, I don't have clue here.

• Cannot start with $n=0$ Feb 27, 2015 at 16:26
• Does it help any to write it as $\sum\frac1n(\frac14)^n$? Does the series $\sum\frac{x^n}{n}$ look familiar at all? Feb 27, 2015 at 16:29

$$\sum_{n=1}^\infty \frac{1}{n2^{2n}}=\sum_{n=1}^\infty \frac{(1/4)^n}{n}=-\ln\left(1-\frac14\right)=-\ln\frac34$$

Since, $$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+...$$

• You are off by a negative and that's because the anti derivative of $\frac{1}{1-x}$ has a negative upfront by the chain rule (+1) Feb 27, 2015 at 16:31
• @imranfat edited Feb 27, 2015 at 16:31
• @AlonAlon If you take the geometric series and integrate (The constant turns out to be zero) and then replace x by 1/4, you get ADG's answer Feb 27, 2015 at 16:32
• @imranfat, following your guidance: $$\sum_{n=0}^\infty \int x^n \, dx = \sum_{n=1}^\infty \frac{x^n}{n}$$. So we have $$\int \frac{1}{1-x} \, dx = -\ln(1-x)$$. Applying $x=4$ we get $-\ln(3/4)$ Feb 27, 2015 at 16:54
• Now, I'm not so sure what should be the bounds of those integrals. Can you help me with that? Feb 27, 2015 at 16:55

We could write this just as well as $$\sum_{n=1}^\infty \frac{1}{n}(1/4)^n$$ Consider the function $$f(x) = \sum_{n=1}^\infty \frac{1}{n}x^n \quad x \in (-1,1)$$ noting that $f(0) = 0$. We evaluate $$f'(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x} \quad x \in (-1,1)$$ It follows that $$f(4) = f(0) + \int_0^{1/4} f'(t)\,dt = \int_0^{1/4} \frac 1{1-t}\,dt$$

• You cannot start with $n=0$ Feb 27, 2015 at 16:33

We have

$$\sum_{n = 1}^\infty \frac{1}{n2^{2n}} = \sum_{n = 1}^\infty \frac{(1/4)^n}{n} = -\log\left(1 - \frac{1}{4}\right) = \log\frac{4}{3}$$

For any $z$ such that $|z|<1$ we have:

$$\sum_{n=1}^{+\infty}\frac{z^n}{n} = \sum_{n=1}^{+\infty}\int_{0}^{z}x^{n-1}\,dx = \int_{0}^{z}\frac{dx}{1-x} = -\log(1-z)$$ and by plugging in $z=\frac{1}{4}$ it follows that:

$$\sum_{n\geq 1}\frac{1}{n 2^{2n}}=\color{red}{\log 4-\log 3}.$$

• You are able to make the interchange because it's a power-series. Right? Feb 27, 2015 at 17:44
• @AlonAlon: We can exchange $\sum$ and $\int$ since we have uniform convergence due to the fact that $|z|<1$. Feb 27, 2015 at 17:50

Think about $\sum_{n=1}^\infty\frac{x^n}{n}=\log{(1-x)}$. The sum of your series is therefore $\log{(4/3)}$