I've deduced two identities involving the Möbius function. After I would like equate both identities, and doing specialization get an identity for which you tell me if both side of the identity reach the same decimal approximation (I have no means to do this last experiement). Then I could to know if all my calculations were rights. I presume that all convergence issues can be proven by absolute convergence for $|x|<1$.

Motivation. In next calculations I get identities involving series representations for simple powers of the inverse sine $\arcsin(x)$ and Lambert series for the Möbius function, by means of derivation, integration and specilizations. The result, if it is feasible, should be identities involving binomial numbers of the shape $\binom{2k}{k}$ and the Möbius function $\mu(n)$. In my thoughts the final statement should be interesting.

Here is my guideline:

Step 1. For $|x|<1$, I've combined the Lambert series for the Möbius function (see this Wikipedia if you need it) with the Generalized Binomial Theorem that provide us the derivative of the inverse sine (that is the first power of this trigonometric function). Multiply by $\frac{1}{4^k}\binom{2k}{k}$ the evaluation of the Lambert series at $x^{2k}$ and after take the sum $\sum_{k\geq 1}$, then you get $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{4^k}\binom{2k}{k}\frac{x^{2nk}}{1-x^{2nk}}=\frac{1}{\sqrt{1-x^2}}.$$ Integrating termwise and doing the evaluation at $\frac{x}{2}$ one has $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{4^k}\binom{2k}{k}\int_0^{\frac{x}{2}}\frac{t^{2nk}}{1-t^{2nk}}dt=\arcsin\left(\frac{x}{2}\right)-0.$$

Step 2. For $|x|<1$, let the expression for the power $\arcsin^2\left(\frac{x}{2}\right)$ that was in the literature, you can deduce from this MathWorld. Combining with the Lambert series as I did in Step 1, one writes $$\frac{1}{2}\sum_{k=1}^\infty\sum_{n=1}^{\infty}\frac{\mu(n)}{\binom{2k}{k}k^2}\frac{x^{2nk}}{1-x^{2nk}}=\arcsin^2\left(\frac{x}{2}\right).$$ We take the derivative termwise, since $\frac{d}{dx}\arcsin^2\left(\frac{x}{2}\right)=\frac{\arcsin\left(\frac{x}{2}\right)}{\sqrt{1-\left(\frac{x}{2}\right)^2}}$ and $\frac{d}{dx}\frac{x^{2nk}}{1-x^{2nk}}=\frac{2nkx^{2nk-1}}{(1-x^{2nk})^2}$, to get


Step 3. For $|x|<1$, one gets by combination of the two deduced identites in previous steps that $$\sqrt{1-\left(\frac{x}{2}\right)^2}\sum_{k=1}^\infty\sum_{n=1}^{\infty}\frac{n\mu(n)}{\binom{2k}{k}k}\frac{x^{2nk-1}}{(1-x^{2nk})^2}=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{4^k}\binom{2k}{k}\int_0^{\frac{x}{2}}\frac{t^{2nk}}{1-t^{2nk}}dt.$$

Background for the question. I would like to verify previous identity, I say the identity of Step 3 for example for $x=\frac{1}{2}$. I know that then the definite integration from $$\int \frac{x^a}{1-x^a}dx=x({_2F_1\left(1 ,\frac{1}{a} ;1+\frac{1}{a};x^{a}\right)}-1)+\text{constant}$$ will be convergent (I've read the conditions from a MathWolrd article, and also I know the arithmetic of Pochhammer symbols). I presume that all my steps can be justified by absolute convergence for $|x|<1$.

Question. Can you provide me a decimal approximation of both sides of the deduced identity in Step 3 at $x=\frac{1}{2}$ to check my calculations (I say that if these are different, then should be a mistake)? Of course if you know that there is a mistake in my calculations or reasoning, especially about the convergence, tell me please. Thanks in advance.

  • $\begingroup$ I am asking if you can do the last step, that is to combine with the formula of the hypergeometric function, and after do an evaluation at $x=1/2$ of both sides of the identity. If when you read this post you recognize a mistake or my claim about the absolute convergence for all manipulations was wrong, please tell me. Many thanks. $\endgroup$ – user243301 Dec 29 '16 at 20:57

You had made ONE mistake at the beginning when you calculate: $$ \sum_{k=\color{red}{1}}^{\infty}\binom{2k}{k}\frac{x^{2k}}{4^k} = \frac{1}{\sqrt{1-x^2}} $$ Actually, $$ \begin{align} \sum_{k=\color{red}{0}}^{\infty}\binom{2k}{k}\left(\frac{x}{2}\right)^{2k} = \sum_{k=\color{red}{0}}^{\infty}\binom{2k}{k}\frac{x^{2k}}{4^k} &= \frac{1}{\sqrt{1-x^2}} \quad\Rightarrow\quad \\[2mm] \sum_{k=\color{red}{1}}^{\infty}\binom{2k}{k}\frac{x^{2k}}{4^k} &= \frac{1}{\sqrt{1-x^2}} \color{red}{-1} \end{align} $$ Thus, $$ \begin{align} \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(n)}{4^k}\binom{2k}{k}\frac{x^{2kn}}{1-x^{2kn}} &= \frac{1}{\sqrt{1-x^2}}-1 \quad\Rightarrow {\small\left\{\text{integrate}\right\}} \\[2mm] \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(n)}{4^k}\binom{2k}{k}\int_0^{\frac{x}{2}}\frac{t^{2nk}}{1-t^{2nk}}dt &= \arcsin\left(\frac{x}{2}\right)-\frac{x}{2} \end{align} $$ The rset is okay, $$ \begin{align} \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(n)}{\binom{2k}{k}}\frac{1}{k^2}\frac{x^{2kn}}{1-x^{2kn}} &= 2\arcsin^2\left(\frac{x}{2}\right) \quad\qquad\Rightarrow {\small\left\{\text{differentiate}\right\}} \\[2mm] \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\mu(n)}{\binom{2k}{k}}\frac{n}{k}\frac{x^{2kn-1}}{(1-x^{2kn})^2} &= \arcsin\left(\frac{x}{2}\right){\LARGE /}\sqrt{1-\left({\small\frac{x}{2}}\right)^2} \end{align} $$ And the numerical calculation at $\,x={\small1/2}\,$ - using Mathematica - support the results:

0.12769462267733195    = NSum[NSum[(MoebiusMu[n]/Binomial[2k,k]) (1/k^2) ((x^(2k n))/(1-x^(2k n))), {k,1,Infinity}], {n,1,Infinity}]
0.12769462267733192    = N[2ArcSin[x/2]^2]
0.2609670453548566     = NSum[NSum[(MoebiusMu[n]/Binomial[2k,k]) (n/k) ((x^(2k n-1))/(1-x^(2k n))^2), {k,1,Infinity}], {n,1,Infinity}]
0.2609670453548564     = N[ArcSin[x/2]/Sqrt[1-(x/2)^2]]
0.1547005383792515     = NSum[NSum[4^(-k) MoebiusMu[n] Binomial[2k,k] ((x^(2k n))/(1-x^(2k n))), {k,1,Infinity}], {n,1,Infinity}]
0.15470053837925168    = N[1/Sqrt[1-x^2]-1]
0.0026802551420786573  = NSum[NSum[4^(-k) MoebiusMu[n] Binomial[2k,k] NIntegrate[(t^(2k n))/(1-t^(2k n)), {t,0,x/2}], {k,1,Infinity}], {n,1,Infinity}]
0.002680255142078647   = N[ArcSin[x/2]-x/2]
  • $\begingroup$ Many thanks then, I work to do a comparison between you calculations and mine. Many thanks for your help. $\endgroup$ – user243301 Feb 12 '17 at 8:51
  • $\begingroup$ U R most welcome. $\endgroup$ – Hazem Orabi Feb 12 '17 at 9:03

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