Some hints to start the evaluation of this series? $$\sum_{k=0}^\infty \dfrac{2^{2k}(k !)^2}{(2k)!(2k+1)^2}\left(\dfrac{1}3-\dfrac{1}{4^{k+1}}\right)$$
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Where did this problem come from? – Mhenni Benghorbal Sep 28 '13 at 08:51
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@MhenniBenghorbal I received it from a kid. I don't know the real source. – Crazy_girl Sep 28 '13 at 08:57
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A kid can deal with kind of sums! – Mhenni Benghorbal Sep 28 '13 at 09:05
3 Answers
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Let's continue Ron Gordon's fine work. He got your sum as : $$I=\frac13 F(1) - \frac12 F\left( \frac12 \right )$$ with $\quad\displaystyle F(x):=\int_0^x \frac{\arcsin{t}}{t\,\sqrt{1-t^2}}dt=\int_0^{\arcsin(x)}\frac u{\sin(u)}du\quad$ (using the substitution $t:=\sin(u)$)
It is easy to show that $\ \displaystyle\log(\tan(u/2))'=\frac 1{\sin(u)}\,$ so that for $\,v:=\tan(u/2)\,$ we get : \begin{align}F(x)&=\int_0^{\arcsin(x)}u\ d(\log(\tan(u/2)))\\ &=\int_0^{\tan(\arcsin(x)/2)}2\arctan(v)\ d(\log(v))\\ &=2\int_0^{\tan(\arcsin(x)/2)}\frac{\arctan(v)}v\ dv\\ &=2\ \rm{Ti}_2\left(\tan\left(\frac{\arcsin(x)}2\right)\right)\\ \end{align}
with $\,\displaystyle\rm{Ti}_2(x):=\int_0^x\frac{\arctan(v)}v\,dv=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^{2k-1}}{(2k-1)^2}\,$ the inverse tangent integral.
Since $\ \arcsin(1)=\dfrac{\pi}2\,$ and $\ \arcsin\left(\frac 12\right)=\dfrac{\pi}6\,$ we may rewrite $I$ as : \begin{align} I&=\frac23 \rm{Ti}_2\left(\tan\left(\frac{\pi}4\right)\right) - \rm{Ti}_2\left(\tan\left(\frac{\pi}{12}\right)\right)\\ &=\frac23 \rm{Ti}_2(1) - \rm{Ti}_2\left(2-\sqrt{3}\right)\\ \end{align} Or simply $$I=-\frac{\pi}{12}\ln(2-\sqrt{3})=\frac{\pi}{12}\ln(2+\sqrt{3})$$
from the identity $(13)$ of the previous MathWorld link (with the incorrect sign corrected to get Lewin's $(2.29)$) which is itself a consequence of the more general identity $(2.20)$ (p.$42$ of Lewin's $1981$ book 'Polylogarithms and Associated Functions') :
$$\rm{Ti}_2\left(\frac{y^2}2\right)+\frac 12\rm{Ti}_2\left(\frac{y(2+y)}{2(1+y)}\right)-\frac 12\rm{Ti}_2\left(\frac{y(2-y)}{2(1-y)}\right)+\rm{Ti}_2\left(\frac{y}{2+y}\right)-\rm{Ti}_2\left(\frac{y}{2-y}\right)+\rm{Ti}_2\left(1-y\right)+\rm{Ti}_2\left(\frac 1{1+y}\right)=2\,\rm{Ti}_2\left(1\right)+\frac{\pi}4\log\left(\frac{1-y}{1+y}\right)$$ applied to the specific case $\,y=\sqrt{3}-1\,$ because $\,y\,$ verifies $\,\dfrac{y\,(2+y)}{2\,(1+y)}=\dfrac1{1+y}$ (Lewin $2.4.2$) .
A more direct derivation was provided by Ramanujan (entry $14$ to $17$ of his first notebook).
To show that $\ \displaystyle \rm{Ti}_2(2-\sqrt{3})=\frac{\pi}{12}\ln(2-\sqrt{3})+\frac 23\rm{Ti}_2(1)\quad$ he used his entry $17$ :
$$\tag{*}\rm{Ti}_2(\tan(x))=x\,\log|\tan(x)|+\sum_{k=0}^\infty \frac{\sin((4k+2)x)}{(2k+1)^2}\quad\text{for}\ |x|<\pi/2$$
This is formula $(3)$ of MathWorld that we may apply to the specific cases $\,\displaystyle x=\frac{\pi}4$ and $\,\displaystyle x=\frac{\pi}{12}$ as indicated by Random Variable.
$\,\displaystyle\sum_{k=0}^\infty \frac{\sin\left((4k+2)\frac{\pi}4\right)}{(2k+1)^2} =K\ $ and $\,\displaystyle\sum_{k=0}^\infty \frac{\sin\left((4k+2)\frac{\pi}{12}\right)}{(2k+1)^2} =\frac 23\,K\quad$ with $K$ the Catalan constant allow to conclude.
The entry $17$ $(*)$ itself was deduced from $\,\displaystyle\rm{Ti}_2(\tan(x))=\frac 12\int_0^{2x}\frac u{\sin(u)}du\ $ and Ramanujan's more general entry $14$ giving a formula for $\,\displaystyle\frac 12\int_0^{x}\frac {u^n}{\sin(u)}du$ (formulae $(14.1)$ to $(14.4)$ from ch.$9$ of Berndt's R.N. Part I).
Raymond Manzoni
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I don't know if it's in Lewin's book, but you can show that $$ \int_{0}^{2-\sqrt{3}} \frac{\arctan x}{x} \ dx = \frac{\pi}{12}\log \left( 2-\sqrt{3}\right) + \sum_{n \text{ odd}}^\infty \frac{\sin \left(\frac{\pi n}{6} \right)}{n^2} $$
The sum can then be evaluated in terms of Catalan's constant.
– Random Variable Sep 28 '13 at 19:56 -
Thanks @RandomVariable. Ramanujan's derivation is indeed more direct and references were added. – Raymond Manzoni Sep 28 '13 at 23:34
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Let
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
Then $f$ has the following series expansion:
$$\sum_{k=0}^{\infty} \frac{2^{2 k}}{\displaystyle(2 k+1)\binom{2k}{k}} x^{2 k+1} $$
so that the sum may be expressed as follows:
$$F(x)=\sum_{k=0}^{\infty} \frac{2^{2 k}}{\displaystyle(2 k+1)^2\binom{2k}{k}} x^{2 k+1} = \int_0^x dt \frac{\arcsin{t}}{t \sqrt{1-t^2}}$$
The sum you seek is then
$$\frac13 F(1) - \frac12 F\left( \frac12 \right )$$
The integral on the RHS of the definition of $F$ may be expressed in terms of a log and a dilogarithm.
EDIT
The actual result is $(\pi/12) \log{(2+\sqrt{3})}$, so this will need further work.
Ron Gordon
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Actually...the actual answer is very very simple! I will be adding to this as soon as I can. – Ron Gordon Sep 28 '13 at 10:50
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Your integral looks simpler with $t:=\sin(x)$ to get $\int \dfrac x{\sin(x)}dx$. – Raymond Manzoni Sep 28 '13 at 11:09
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@RaymondManzoni: doesn't help much. I'm on my way to the solution. Stay tuned. – Ron Gordon Sep 28 '13 at 11:16
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In fact it helps since alpha while for $\pi/2$ you get $2C$ (the Catalan $C$ cancel). – Raymond Manzoni Sep 28 '13 at 11:17
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Having trouble with the $\pi/6$ piece and am out of time. So we will have to wait; perhaps someone will fill in my gap. – Ron Gordon Sep 28 '13 at 12:07
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That's interesting.
You have Catalan's constant for both parts of your series.
$$\sum_{k=0}^\infty \dfrac{2^{2k}(k!)^2}{(2k)!(2k+1)^2}=2\mathrm G$$
you can find it here (Entry 22).
And
$$\sum_{k=0}^\infty \dfrac{2^{2k}(k
!)^2}{(2k)!(2k+1)^2}\dfrac{1}{4^{k+1}}=\dfrac{1}4\sum_{k=0}^\infty \dfrac{(k
!)^2}{(2k)!(2k+1)^2}= \frac{2 \mathrm G}{3}-\frac{1}{12} \pi \log \left(2+\sqrt{3}\right)$$
you can find here.
So
$$\sum_{k=0}^\infty \dfrac{2^{2k}(k
!)^2}{(2k)!(2k+1)^2}\left(\dfrac{1}3-\dfrac{1}{4^{k+1}}\right)=\frac{\pi}{12} \log \left(2+\sqrt{3}\right)$$
Caran-d'Ache
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