1
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ Cannot start with $n=0$ $\endgroup$
    – marwalix
    Feb 27, 2015 at 16:26
  • $\begingroup$ Does it help any to write it as $\sum\frac1n(\frac14)^n$? Does the series $\sum\frac{x^n}{n}$ look familiar at all? $\endgroup$ Feb 27, 2015 at 16:29

5 Answers 5

7
$\begingroup$

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


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

$\endgroup$
8
  • $\begingroup$ 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) $\endgroup$
    – imranfat
    Feb 27, 2015 at 16:31
  • $\begingroup$ @imranfat edited $\endgroup$
    – RE60K
    Feb 27, 2015 at 16:31
  • 1
    $\begingroup$ @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 $\endgroup$
    – imranfat
    Feb 27, 2015 at 16:32
  • $\begingroup$ @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 $-\ln(3/4)$ $\endgroup$
    – AlonAlon
    Feb 27, 2015 at 16:54
  • $\begingroup$ Now, I'm not so sure what should be the bounds of those integrals. Can you help me with that? $\endgroup$
    – AlonAlon
    Feb 27, 2015 at 16:55
4
$\begingroup$

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 $$ f(4) = f(0) + \int_0^{1/4} f'(t)\,dt = \int_0^{1/4} \frac 1{1-t}\,dt $$

$\endgroup$
1
  • 1
    $\begingroup$ You cannot start with $n=0$ $\endgroup$
    – marwalix
    Feb 27, 2015 at 16:33
3
$\begingroup$

We have

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

$\endgroup$
3
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ You are able to make the interchange because it's a power-series. Right? $\endgroup$
    – AlonAlon
    Feb 27, 2015 at 17:44
  • 1
    $\begingroup$ @AlonAlon: We can exchange $\sum$ and $\int$ since we have uniform convergence due to the fact that $|z|<1$. $\endgroup$ Feb 27, 2015 at 17:50
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.