25
$\begingroup$

show that

$$\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}=5+4\sqrt{2}\left(\log{\dfrac{2\sqrt{2}}{1+\sqrt{2}}}-1\right)$$

where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$

My try: we let $$s(x)=\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{(n+1)}x^{n+1}$$ then $$s'(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\binom{2n}{n}H_{n+1}x^n$$ then I can't.Thank you

$\endgroup$
  • 1
    $\begingroup$ I'd start with the Catalan numbers, which are $\frac{1}{n+1}\binom{2n}{n}$. Not sure how much that would help, but it seems important. en.wikipedia.org/wiki/Catalan_number $\endgroup$ – Thomas Andrews Mar 29 '14 at 2:50
  • $\begingroup$ The other possibility is to represent the $H_n$ in $\Sigma$ notation and swap the order of summation, but that seems dangerous because you'd need some absolute convergence condition. $\endgroup$ – Thomas Andrews Mar 29 '14 at 2:53
  • $\begingroup$ Not sure how helpful this would be, but I asked a question some time ago that had to do with harmonic series... maybe it could help? math.stackexchange.com/questions/681296/… $\endgroup$ – 2012ssohn Mar 29 '14 at 3:10
29
+25
$\begingroup$

There's a transformation formula (might have been derived from generalized Euler series transformation): $$ f(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, a_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, g\left(\dfrac{x}{1+4\, x}\right) $$ where $$a_n=\sum_{k=0}^n\, \binom{n}{k}\, (-1)^{n-k}\, b_k$$ and $$g(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, b_n \, x^n$$

For $a_n=(-1)^{n-1}\, H_n$ and $b_n=\dfrac{1}{n}$, we have the identity $$ (-1)^{n-1}\, H_n = \sum_{k=1}^n \binom{n}{k}\, \dfrac{(-1)^{n-k}}{n} $$

Also, from the generating function for central binomial coefficients, the following g.f. can be obtained: $$ G_1(x)=\sum_{n=1}^\infty \, \binom{2n}{n}\, \dfrac{1}{n}\, x^n = 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, x}}\right)} $$

Therefore,

$\displaystyle \begin{aligned} f(x)&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, H_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, \sum_{n=1}^\infty \, \binom{2n}{n} \, \dfrac{1}{n}\, \left( \dfrac{x}{1+4\, x} \right)^n\\\\ &= \dfrac{1}{\sqrt{1+4\, x}}\, G_1\left(\dfrac{x}{1+4\, x}\right)\\\\ &= \dfrac{1}{\sqrt{1+4\, x}}\, 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, \left(\dfrac{x}{1+4\, x}\right)}}\right)} \end{aligned} $

Then, consider the g.f. we need in order to get the required sum:

$\displaystyle \begin{aligned} h(x) &=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n+1}}{n+1} \, x^n\\\\ &=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}+\dfrac{1}{n+1}}{n+1} \, x^n\\\\ &=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}}{n+1} \, x^n+\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{1}{(n+1)^2} \, x^n\\\\ &=h_1(x)+h_2(x) \end{aligned} $

Both $h_1(x)$ and $h_2(x)$ can be found by integrating the g.f.'s we have already seen:

$\displaystyle \begin{aligned} h_1(x)&=\dfrac{1}{x}\int f(x) \, dx+\dfrac{4 \, \log\left(2\right) + 1}{4 \, x}\\\\ &=\dfrac{4 \, \sqrt{4 \, x + 1} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{4 \, x + 1}} + 1}\right) - 4 \, \log\left(\sqrt{4 \, x + 1} + 1\right) - 1}{4 \, x} + \dfrac{4 \, \log\left(2\right) + 1}{4 \, x} \end{aligned}$

$\displaystyle \begin{aligned} h_2(x)&= \dfrac{1}{x}\, \int \left(\dfrac{1}{x}\, \int -\dfrac{1}{\sqrt{1+4\, x}}\, dx \right)\, dx + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1\\\\ &= -\dfrac{2 \, \sqrt{4 \, x + 1} - \log\left(\sqrt{4 \, x + 1} + 1\right) + \log\left(\sqrt{4 \, x + 1} - 1\right)}{2 \, x} + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1 \end{aligned} $

and the required sum is: $\displaystyle \begin{aligned} h\left(\dfrac{1}{4}\right)&=4 \, \sqrt{2} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{2}} + 1}\right) - 4 \, \sqrt{2} + 4 \, \log\left(2\right) + 2 \, \log\left(\dfrac{1}{4}\right) - 2 \, \log\left(\sqrt{2} + 1\right) - 2 \, \log\left(\sqrt{2} - 1\right) + 5\\\\ &=4 \, \sqrt{2} {\left(\log\left(\dfrac{2 \, \sqrt{2}}{\sqrt{2} + 1}\right) - 1\right)} + 5\\\\ &\approx 0.238892690197059 \end{aligned}$

References:

[1]

[2]

$\endgroup$
12
$\begingroup$

Note: This is not an independent answer. It's only a supplement to the very interesting answer presented from gar. When going through the details I wrote some notes for me. Maybe they are also of interest for other readers.


The proof of the identity

$$\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)} =5+4\sqrt{2}\left(\log\frac{2\sqrt{2}}{1+\sqrt{2}}-1\right) $$

is based upon four ingredients, which are cleverly combined. It takes one binomial inverse pair, one identity of the Harmonic numbers, the generating functions for central binomials and most important the powerful series transformation formula from Euler, which is the key of the proof.

We start with a binomial inverse pair. To show the relationship we multiply exponential generating functions (egfs). Let $A(x)=\sum_{n\ge0}a_{n}\frac{x^n}{n!}$ and $B(x)=\sum_{n\ge0}b_{n}\frac{x^n}{n!}$ egfs with $B(x)=A(x)e^x$. Comparing coefficients gives the following

Binomial inverse pair \begin{align} B(x)&=A(x)e^x&A(x)&=B(x)e^{-x}\\ b_n&=\sum_{k=0}^{n}\binom{n}{k}a_k&a_n&=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}b_k\qquad (1) \end{align}

This simple one and many other examples of binomial inverse pairs can be found e.g. in Combinatorial Identities a classic from John Riordan ($1968$).

There we can also find an easy proof of the next ingredient (ch.$1$, Example $3$):

Identity of Harmonic Numbers $H_n$ \begin{align} H_n&=\sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}\frac{1}{k} \end{align}

Let \begin{align} f_n&=\sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}\frac{1}{k} =\sum_{k=1}^{n}(-1)^{k+1}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)\frac{1}{k}\\ &=f_{n-1}-\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}=f_{n-1}-\frac{1}{n}\left((1-1)^n-1\right)\\ &=f_{n-1}+\frac{1}{n}\\ &=H_n \end{align} Now we consider $(-1)^{n-1}H_n$, observe that it matches with the binomial pairs $(1)$ above and get an

Inverse pair with Harmonic Numbers $H_n$ \begin{align} (-1)^{n-1}H_n=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{n-k}}{k}&&\frac{1}{n}=\sum_{k=1}^{n}\binom{n}{k}(-1)^{k-1}H_k\qquad(2) \end{align}

The third ingredient and most important for the proof is one of the treasures from the master:

Euler's series transformation formula:

Given a function $f(x)=\sum_{n\ge0}a_nx^n$ analytical on the unit disk, the following representation is valid: \begin{align} f(x)&=\sum_{n\ge0}a_nx^n\\ \frac{1}{1-x}f\left(\frac{x}{1-x}\right)&=\sum_{n\ge0}\left(\sum_{k=0}^{n}\binom{n}{k}a_k\right)x^k\qquad(3) \end{align}

Observe that the transformation of the coefficients $a_n$ to $\sum_{k=0}^{n}\binom{n}{k}a_k$ which is usually done via egfs (see above) is here presented via ordinary generating functions (ogfs) and with interesting consequences. This transformation formula together with a proof using Cauchy's integral formula can be found in Harmonic Number Identities Via Euler's transform from Boyadzhiev ($2009$).

His proof: Using Cauchy's integral formula and given a function $f$ as above, we have $a_k=\frac{1}{2\pi i}\oint_{L}\frac{1}{\lambda^k}\frac{f(\lambda)}{\lambda}d\lambda$ for an appropriate closed curve $L$ around the origin. Multiplying both sides by $\binom{n}{k}$ and summing for $k$ we find \begin{align} \sum_{k=n}^{n}\binom{n}{k}a_k&=\frac{1}{2\pi i}\oint_{L}\left(\sum_{k=0}^{n}\binom{n}{k}\frac{1}{\lambda^k}\right)\frac{f(\lambda)}{\lambda}d\lambda=\frac{1}{2\pi i}\oint_{L}\left(1+\frac{1}{\lambda}\right)^n\frac{f(\lambda)}{\lambda}d\lambda \end{align} Multiplying this by $x^n$ ($x$ small enough) and summing for $n$ we arrive at the desired representation, because

$$\sum_{n=0}^{\infty}x^n\left(1+\frac{1}{\lambda}\right)^n=\frac{1}{1-x(1+\frac{1}{\lambda})} =\frac{1}{1-x}\frac{\lambda}{\lambda-\frac{x}{1-x}}$$

Note: A simple pure formal proof of a variant of this transformation is presented in Transformations to Speed the Convergence of Series by Rosser ($1951$).

In another paper Series transformation formulas of Euler type ($2009$) Boyadzhiev gives us with the substitution $\frac{x}{1-x}=z$ and introducing some extensions of the transformation formula $(3)$ following variation:

For any complex number $\alpha$ the following representation is true: $$\sum_{n=0}^{\infty}\binom{\alpha}{n}(-1)^na_nz^n =(z+1)^\alpha\sum_{n=0}^{\infty}\binom{\alpha}{n}(-1)^n\sum_{k=0}^{n}\binom{n}{k}a_k\left(\frac{z}{z+1}\right)^n\qquad (4) $$

Now, formula $(4)$ is used by Boyadhziev in his paper Series with Central Binomial Coefficients, Catalan Numbers, and Harmonic Numbers ($2012$) to obtain by setting $\alpha=-\frac{1}{2}$ and noting that $\binom{-1/2}{n}=\frac{(-1)^n}{4^n}\binom{2n}{n}$

$$\sum_{n=0}^{\infty}\binom{2n}{n}a_n\frac{z^n}{4^n} =\frac{1}{\sqrt{1+z}}\sum_{n=0}^{\infty}\binom{2n}{n}\sum_{k=0}^{n}\binom{n}{k}a_k\frac{1}{4^n}\left(\frac{z}{z+1}\right)^n\qquad (5) $$

Setting $z=4x$, $a_n=(-1)^{n-1}H_n$ and observing the binomial inverse of $(-1)^{n-1}H_n$ from $(2)$ we get

$$\sum_{n=0}^{\infty}\binom{2n}{n}(-1)^{n-1}H_{n}x^n =\frac{1}{\sqrt{1+4x}}\sum_{n=1}^{\infty}\binom{2n}{n}\frac{1}{n}\left(\frac{x}{1+4x}\right)^n\qquad (6) $$

This was the heavy part. The fourth and last ingredient is the well known

Series expansion for the central binomial coefficient: $$\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$

Checking the RHS of (6) we see, that we need an additional factor $\frac{1}{n}$. To switch from $a_n$ to $\frac{1}{n}a_n$ we consider the ogf $A(x)=\sum_{n\ge 1}a_{n}x^{n}$ and observe

\begin{align} &(a_n)_{n\ge 1}&&\left(\frac{1}{n}a_n\right)_{n\ge 1}\\ A(x)&=\sum_{n\ge 1}a_{n}x^{n}&\int_{0}^{x}&\frac{1}{t}A(t)dt=\sum_{n\ge 1}\frac{a_n}{n}x^n \end{align}

Applying this transformation to the ogf of the central binomial coefficient gives

\begin{align} \sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}&=\int_{0}^{x}\frac{1}{t}\left(\frac{1}{\sqrt{1-4t}}-1\right)dt\\ &=2 \log\left(\frac{2}{1+\sqrt{1-4x}}\right)\qquad\qquad(7) \end{align}

Finally, combining $(6)$ and $(7)$ results in \begin{align} \sum_{n=0}^{\infty}&\binom{2n}{n}(-1)^{n-1}H_{n}x^n\\ &=\frac{2}{\sqrt{1+4x}}\log\left(\frac{2}{1+\sqrt{1-4\left(\frac{x}{1+4x}\right)}}\right)\\ &=f(x) \end{align}

with $f(x)$ the function used in the answer of gar. The series is also converging for $x=-\frac{1}{4}$.

All following calculations are clearly and completely shown in the answer from gar.

$\endgroup$
  • 1
    $\begingroup$ Hi Markus, thanks for providing that. $\endgroup$ – gar Apr 6 '14 at 12:04
  • $\begingroup$ My pleasure! I could repeat a lot and I've learned a lot. $\endgroup$ – Markus Scheuer Apr 6 '14 at 12:16
5
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{2n \choose n}{\pars{-1}^{n - 1}H_{n + 1} \over 4^{n}\pars{n + 1}} = 5 + 4\root{2}\bracks{\ln\pars{2\sqrt{2} \over 1 + \root{2}} - 1} \approx {\tt 0.2389}:\ {\large ?}}$

\begin{align} \mbox{Note that}\quad\sum_{n = 1}^{\infty} {2n \choose n}{\pars{-1}^{n - 1}H_{n + 1} \over 4^{n}\pars{n + 1}}= -\sum_{n = 1}^{\infty}{2n \choose n}{H_{n + 1} \over n + 1}\, \pars{-\,{1 \over 4}}^{n}\tag{1} \end{align}

In a previous answer, I did a detailed derivation of: $$ \sum_{n = 1}^{\infty}{2n \choose n}{H_{n + 1} \over n + 1}\,\mu^{n} =-1 - \int_{0}^{1}{\ln\pars{1 - t} \over \root{1 - 4\mu t}}\,\dd t $$

such that $\pars{1}$ is reduced to: \begin{align}&\color{#66f}{\large\sum_{n = 1}^{\infty} {2n \choose n}{\pars{-1}^{n - 1}H_{n + 1} \over 4^{n}\pars{n + 1}}}= 1 +\ \overbrace{\int_{0}^{1}{\ln\pars{1 - t} \over \root{1 + t}}\,\dd t} ^{\ds{\mbox{Set}\ x \equiv \root{1 + t}}} \\[3mm]&=1 + 2\int_{1}^{\root{2}} \bracks{\ln\pars{\root{2} + x} + \ln\pars{\root{2} - x}}\,\dd x \\[3mm]&=\color{#66f}{\large% 5 + 4\root{2}\bracks{\ln\pars{2\sqrt{2} \over 1 + \root{2}} - 1}} \approx {\tt 0.2389} \end{align}

$\endgroup$

Your Answer

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