7
$\begingroup$

We all know the series expansion $$\log 2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$ I also am able to use the method of Wilf-Zeilberger to the effect that $$\log 2=3\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\binom{2n}n2^n}. \tag2$$

QUESTION. Can you provide yet another proof of the formula in (2)?

Remark. My motivation for this question goes beyond this particular series, hoping it paves a way forward in my study.

Postscript. After those generous replies (see below), it appears that the idea rests on $$\log\left(1+\frac1x\right)=2\sinh^{-1}\left(\frac1{2\sqrt{x+x^2}}\right)$$ so that we may put $x=1$ to obtain (1) and (2).

To reveal the background: (2) is found from (1) by a "series acceleration" method which does not even stop there. In fact, stare at these two \begin{align*}\log 2&=3\sum_{n=1}^{\infty}\frac{14n-3}{\binom{2n}2\binom{4n}{2n}2^{2n+1}}, \tag3 \\ \log 2&=3\sum_{n=1}^{\infty} \frac{(171n^2 - 111n + 14)(-1)^{n-1}}{\binom{3n}3\binom{6n}{3n}2^{3n+1}} \tag4 \end{align*} One may now ask: can you furnish an alternative proof for the formulae (3) or (4)?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Your reference gives formulas for series involving zeta functions. Can you point out anything about (2) in the reference? $\endgroup$
    – T. Amdeberhan
    Commented Jul 17, 2021 at 19:37
  • 1
    $\begingroup$ This is classical result due to Lehmer: Interesting Series Involving the Central Binomial Coefficient, D. H. Lehmer, The American Mathematical Monthly, Vol. 92, No. 7 (1985), pp. 449-457. $\endgroup$
    – Nemo
    Commented Jul 18, 2021 at 8:26
  • $\begingroup$ these $\log 2$ formulas remind me of the binomial sums for $\pi$, such as $\pi = \sum_{n=0}^\infty \dfrac{50n-6}{{\displaystyle \tbinom{3n}{n}2^{n}}}$ $\endgroup$
    – Carlo Beenakker
    Commented Jul 18, 2021 at 19:51
  • $\begingroup$ It looks like for each $k$ there is a unique similar polynomial with rational coefficients for the denominator $\binom{kn}{k}\binom{2kn}{kn}2^{kn}$ (and alternating signs for odd $k$), e.g. for $k=4$ $$\log 2=\frac14\sum_{n=1}^{\infty} \frac{(14560 n^3 - 16176 n^2 + 5138 n - 417) }{\binom{4n}4\binom{8n}{4n}2^{4n}} $$ or for $k=5$: $$\log 2=\frac1{40}\sum_{n=1}^{\infty} \frac{(2275625n^4 - 3615750n^3 + 1935175n^2 - 395010n + 23544)(-1)^{n-1} }{\binom{5n}5\binom{10n}{5n}2^{5n}} $$ $\endgroup$
    – Wolfgang
    Commented Nov 29, 2023 at 14:47

3 Answers 3

Reset to default
15
$\begingroup$

Since you wish to develop techniques, you might want to consider the more general form $$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k\binom{2n}n2^n}.$$ The arcsine representation $$\arcsin^2z=\frac12\sum_{n=1}^\infty\frac{(2z)^{2n}}{n^2{2n \choose n}}$$ directly gives $$S_2=\tfrac{1}{2}\ln^2 2,$$ (substitute $z=2^{-3/2}i$), upon differentiation one finds $$S_1=\tfrac{1}{3}\ln 2,$$ $$S_0=\tfrac{1}{9}+\tfrac{4}{27}\ln 2,$$ and upon integration, $$S_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$ $$S_4=4\operatorname{Li}_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$ This method apparently fails to give a closed form expression for $k>4$, see this MSE posting.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is another valuable/informative point. Thanks. Upvoted. $\endgroup$
    – T. Amdeberhan
    Commented Jul 17, 2021 at 19:56
13
$\begingroup$

Write the $n$-th term, $(-1)^{n-1} \!\left/ \bigl(n {2n \choose n} 2^n\bigr) \right.$, as the definite integral $$ \frac14 \int_0^1 \left(-\,\frac{x-x^2}{2} \right)^{n-1} dx $$ using the formula for the Beta integral $B(n,n) = \int_0^1 (x-x^2)^{n-1} dx$. Thus the sum over $n$ is $$ \frac14 \int_0^1 \left( 1 + \frac{x-x^2}{2} \right)^{\!-1} dx = \frac12 \int_0^1 \frac{dx}{2+x-x^2} \, dx, $$ which is elementary: the denominator factors as $(1+x)(2-x)$, so expand the integrand in partial fractions and integrate termwise, obtaining $$ \frac16 \int_0^1 \left(\frac1{x+1} + \frac1{2-x}\right) dx = \frac16 (\log 2 + \log 2) = \frac13 \log 2, $$ QED

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ this is cute. Thanks. I wish to see more techniques from others. Upvoted. $\endgroup$
    – T. Amdeberhan
    Commented Jul 17, 2021 at 19:49
10
$\begingroup$

It is well known that $$ 2\left(\sin^{-1}\frac{\sqrt{x}}{2}\right)^2 = \sum_{n\geq 1} \frac{x^n}{n^2{2n\choose n}}. $$ See e.g. here or Enumerative Combinatorics, vol. 1, second ed., Exercise 1.173. Differentiate with respect to $x$, put $x=-1/2$, and use $$ \sin^{-1}z = -i\log(iz+\sqrt{1-z^2}) $$ to deduce (2).

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is equally cute. Thanks! Upvoted. $\endgroup$
    – T. Amdeberhan
    Commented Jul 17, 2021 at 19:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .