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I have come across the following series, which I suspect converges to $\ln 2$:

$$\sum_{k=1}^\infty \frac{1}{4^k(2k)}\binom{2k}{k}.$$

I could not derive this series from some of the standard expressions for $\ln 2$. The sum of the first $100 000$ terms agrees with $\ln 2$ only up to two digits.

Does the series converge to $\ln 2$?

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    $\begingroup$ WA says it's true. $\endgroup$ – lhf Jul 1 '14 at 13:43
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    $\begingroup$ Good to know. How can I derive a proof of it from that link? $\endgroup$ – Pierre-Guy Plamondon Jul 1 '14 at 13:45
  • $\begingroup$ @Pierre-Guy Plamondon Does not replace the analytical answers given below but you could have got much closer to the true value than 2 digits. I got .693147294898206 summing only $2^{16}$ terms. Notice it disagrees with $\ln 2$ in the 7th decimal place! $\endgroup$ – bobbym Jul 1 '14 at 14:21
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Hint: (or outline -- a lot of details and justifications must be done where there are $(\star)$'s)

  • for $\lvert x\rvert < \frac{1}{4}$, \begin{align*} f(x)&\stackrel{\rm def}{=}\frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\frac{x^k}{k} = \frac{1}{2}\sum_{k=1}^{\infty} \binom{2k}{k}\int_0^x t^{k-1}dt \\ &\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k-1}\right)dt = \frac{1}{2} \int_0^x \frac{1}{t} \left(\sum_{k=1}^{\infty} \binom{2k}{k} t^{k}\right)dt \\ &\stackrel{(\star)}{=} \frac{1}{2} \int_0^x \frac{1}{t} \left(\frac{1}{\sqrt{1-4t}}-1\right)dt \\ &= \frac{1}{2} 2\ln\frac{2}{\sqrt{1-4x}+1} \end{align*}
  • "so" ($\star$) \begin{align*} f(x)\xrightarrow[t\to\frac{1}{4}^-]{}\ln 2 \tag{$\star$} \end{align*}
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  • $\begingroup$ A few more comments: the first star is not hard (power series within the radius of convergence: swapping integral and sum is justified); the second is a possibly technical but not conceptually challenging computation of integral (I cheated with Wolfram Alpha). Finally, if I'm not mistaken, for the last step you'll need to invoke Tauber's Theorem; and in particular show that (it is not hard with Stirling's approximation) $\binom{2k}{k}\frac{1}{k4^k} = o\!\left(\frac{1}{k}\right)$. $\endgroup$ – Clement C. Jul 1 '14 at 13:50
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Hint: $\displaystyle\sum_{k=0}^\infty\binom{2k}kx^k$ is nothing else than the binomial series of $~\dfrac1{\sqrt{1-4x}}$ . Now integrate both sides with regard to x, and then let $x=\dfrac14$ .

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Recall that for $|x|<1$ we have $$\frac{1}{\sqrt{1-x^{2}}}=\sum_{k=0}^{\infty}{2n \choose n}\frac{x^{2n}}{4^{n}}$$ shifting over the first terms on the series and dividing by x gives $$\sum_{n=1}^{\infty}{2n\choose n}\frac{x^{2n-1}}{4^{n}}= \frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\frac{1}{x}$$ Now given an small $\varepsilon>0$ we can integrate both sides from $0$ to $1-\varepsilon$, the swap on the RHS is justified due to the uniform convergence, next we let $\varepsilon$ shrink to zero. Wallis product formula yields $$\frac{1}{4^{n}2n}{2n \choose n}\sim\frac{1}{\sqrt{2\pi}}\frac{1}{n\sqrt{2n+1}}$$ which allows us to move in the limit inside and letting $\varepsilon \rightarrow 0$ Hence we have that $$\sum_{n=1}^{\infty}{2n\choose n}\frac{1}{4^{n}2n}= \int_{0}^{1}{\frac{1-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\frac{dx}{x}}=\int_{0}^{\pi/2}\frac{1-\cos(u)}{\sin(u)}du=\int_{0}^{\pi/2}\frac{\sin(u/2)}{\cos(u/2)}du$$

It is pretty obvious what to do next..

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