# Computing $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n}$ with a power series- What did I do wrong?

The requested sum: $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n}= \frac{1}{2}\sum_{0}^{\infty}\frac{1}{4^nn}$

My helper- this power series: $\sum_{0}^{\infty}\frac{x^{n}}{4^n}=\frac{1}{1-\frac{4}{x}}$

Integration due to uniform convergence: $\int \sum_{0}^{\infty}\frac{x^{n}}{4^n}=\int \frac{1}{1-\frac{4}{x}}$

This is what I get:$\sum_{0}^{\infty}\frac{x^{n+1}}{4^{n+1}(n+1)}=-4\ln(4-x)$

Or: $\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-4\ln(4-x)$

Finally what we need is: $\frac{1}{2}\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-2\ln(4-x)$

Now I want to plug-in $x=1$ and get the requested result, but what bothers me is that this is a positive series and I get a negative number in the RHS, obivously something's wrong.

Please tell me what's wrong with the steps described above.

Thanks! :)

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For starters, $\sum_{n = 0}^{\infty} x^n/4^n = \frac{1}{1-(x/4)}$, not $\frac{1}{1-(4/x)}$. – Martin Wanvik Feb 13 '12 at 16:19
Well, first, the sum on the right hand side of the first line should start at $n=1$, not $n=0$. – David Mitra Feb 13 '12 at 16:23
@MartinWanvik: This is a type mistake also. the integral is correct. – Jozef Feb 13 '12 at 16:32
@DavidMitra: sorry, this is a type mistake. Still I don't understand how come it is negative. – Jozef Feb 13 '12 at 16:33
When integrating, you need definite integrals$$\sum_{n=1}^\infty \int_0^x (t/4)^n \,dt = \int_0^x {(t/4)\over 1-(t/4) }\, dt$$ – David Mitra Feb 13 '12 at 16:34

The problem with your method is that you're using primitives and not definite integrals:

$$\sum\limits_{n = 0}^\infty {\frac{1}{{n + 1}}{{\left( {\frac{x}{4}} \right)}^{n + 1}}} = - \log \left( {1 - \frac{x}{4}} \right) = - \log \left( {4 - x} \right) + \log 4 = \int\limits_0^x {\frac{1}{{4 - t}}dt}$$

Since you know

$$-\log \left( {1 - x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}}$$

You simply plug in $1/4$. You get

\eqalign{ & - \frac{1}{2}\log \left( {1 - \frac{1}{4}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr & \frac{1}{2}\log \left( {\frac{4}{3}} \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{n{4^n}}}} \cr}

You should be thinking about differentiating, not integrating. You have

$$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}}$$

So you might want to find

$$\frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} = f\left( x \right)$$

Adn the plug in $1/4$. Use

\eqalign{ & F\left( x \right) = \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \cr & x\frac{d}{{dx}}F\left( x \right) = x\frac{d}{{dx}}\frac{1}{{1 - x}} = x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr & xF'\left( x \right) = \frac{x}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 1}^\infty {n{x^n}} \cr & \frac{1}{2}xF'\left( x \right) = \frac{1}{2}\frac{x}{{{{\left( {1 - x} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{x^n}} \cr}

Now plug in $1/4$ to get

$$\frac{1}{2}\frac{{\frac{1}{4}}}{{{{\left( {1 - \frac{1}{4}} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}}$$

$$\frac{2}{9} = \frac{1}{2}\sum\limits_{n = 0}^\infty {n{{\left( {\frac{1}{4}} \right)}^n}}$$

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Thanks Peter, but this is not the requested sum :) – Jozef Feb 13 '12 at 16:28
@Jozef (!) Sorry, I misread the $n$! Will ad the right answer shortly. – Pedro Tamaroff Feb 13 '12 at 16:30
@Jozef Ok, what about now? – Pedro Tamaroff Feb 13 '12 at 16:33
but it suppose to work in all the radius of convergence which is $(-4,4)$, so what's wrong with $1$, how come it is negative? something's wrong with my way? – Jozef Feb 13 '12 at 16:36
You're using primitves instead of definite integrals. You have a constant $C$ you're not considering. – Pedro Tamaroff Feb 13 '12 at 16:46

It would be best to proceed as Peter does starting with the series representation of $\ln(1-x)$.

You wish to compute $$\sum_{n=1}^\infty {1\over 2^{2n+1} n}= {1\over2} \sum_{n=1}^\infty { 1\over n \, 4^n }$$ (this was your first error, the sum starts at $n=1$).

Using the Geometric series: $$\tag{1} \sum_{n=1}^\infty (x/4)^n= {x/4\over 1-(x/4)}= {1\over 1-(x/4)}-1$$ (your sum of the series was your second error).

If $|x|<4$: $$\sum_{n=1}^\infty\int_0^x (t/4)^n\,dt= \int_0^x {1\over 1-(t/4)}-1\, dt$$ (note, you need to take definite integrals).

Whence $$\sum_{n=1}^\infty{x^{n+1}\over 4^n (n+1)}= -4\ln|1-(x/4)|-x$$

Substituting $x=1$ gives: $$\sum_{n=1}^\infty{1\over 4^n (n+1)}= -4\ln(3/4)-1\approx0.15073$$

So \eqalign{ {1\over2} \sum_{n=1}^\infty { 1\over n \, 4^n } &={1\over2} \sum_{n=0}^\infty { 1\over( n +1)\, 4^{n+1} }\cr &={1\over8} \sum_{n=0}^\infty { 1\over( n +1)\, 4^{n } }\cr &={1\over8} [{1} -4\ln(3/4)-1]\cr &={1\over8} [ -4\ln(3/4) ]\cr &= { \ln(4/3)\over 2 }\cr }

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David, Thanks a lot! – Jozef Feb 13 '12 at 17:35