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I am in doubt with this question .

let $S=\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+...$ Is it converge to $\ln 2$ ?

I tried this $$x=\dfrac12 \to 1+x+x^2+x^3+x^4+...\sim\dfrac{1}{1-x}\to 2$$ by integration wrt x we have $$\int (1+x+x^2+x^3+x^4+...)dx=\int (\dfrac{1}{1-x})dx \to\\ x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+...=-\ln(1-x)$$then put $x=0.5$ $$\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+..\sim -\ln(0.5)=\ln 2$$ now my question is : Is my work true ?
I am thankful for you hint,guide,idea or solutions. (I forgot some technics of calculus)

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    $\begingroup$ Looks good to me! :) $\endgroup$ – Juanito May 22 '17 at 6:32
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    $\begingroup$ It is amazing that rewriting $$S=\dfrac12 +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+...=\dfrac{(\dfrac12)^1}{1}+\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+...$$ makes the problem nicer. $\endgroup$ – Claude Leibovici May 22 '17 at 7:16
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For completeness, you shoud show that the series for $1/(1-x)$ is uniformly convergent in $[0,1/2]$.

As

$$\left|S_n-\frac1{1-x}\right|=\left|\frac{1-x^{n+1}}{1-x}-\frac1{1-x}\right|=\left|\frac{x^{n+1}}{1-x}\right|\le\left|\frac1{2^n}\right|$$ this is ensured and you can integrate term-wise.

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Yes.

This series was already known to Jacob Bernoulli (Gourdon and Sebah http://plouffe.fr/simon/articles/log2.pdf, formula 14) and can be written

$$S=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}$$

To evaluate it, we can change the identity

$$\int_0^1 x^n dx = \frac{1}{n+1}$$

into

$$\int_0^\frac{1}{2} x^n dx = \frac{1}{n+1}\left(\frac{1}{2}\right)^{n+1}$$

and then

$$\begin{align} S&=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}\\ &=\sum_{k=0}^\infty \int_0^\frac{1}{2} x^k dx \\ &=\int_0^\frac{1}{2}\left(\sum_{k=0}^\infty x^k\right) dx\\ &=\int_0^\frac{1}{2} \frac{1}{1-x} dx\\ &=-\log(1-x)|_0^\frac{1}{2}\\ &=\log(2) \\ \end{align}$$

A nice way to encode formulas like

$$\log(2)=\sum_{n=1}^\infty \frac{1}{n2^n}$$ is noting that the numerator is one so the sequence of integer denominators can represent the series. When we search the OEIS for $2,8,24,64$ (http://oeis.org/A036289) we find that

$$\sum_{n=1}^\infty \frac{1}{a(n)} = \log(2)$$ is one of the formulas given.

Your series is a base-2 BBP-type formula for $\log(2)$. The base-3 version is

$$\log(2)=\frac{2}{3} \sum_{k=0}^\infty \frac{1}{(2k+1)9^k} $$

and the sequence of denominators is OEIS http://oeis.org/A155988.

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Your work is good, but it can be better justified with the theory of power series.

The given series is an instance of the power series $$ f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k} $$ for $x=1/2$. The power series has convergence radius $1$; indeed, the ratio test gives $$ \left|\frac{x^{k+1}/(k+1)}{x^k/k}\right|=\frac{k}{k+1}|x|\to |x| $$ Thus you know your series converges for $|x|<1$.

The function $f$, defined over $(-1,1)$, is differentiable and $$ f'(x)=\sum_{k=1}^{\infty}x^{k-1}=\sum_{k=0}^{\infty} x^k=\frac{1}{1-x} $$ (geometric series). Since $f(0)=0$, you can conclude that, for $x\in(-1,1)$, $$ f(x)=\int_0^x\frac{1}{1-t}\,dt=-\ln(1-x) $$ Therefore $$ f(1/2)=-\ln\Bigl(1-\frac{1}{2}\Bigr)=\ln2 $$

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