For integers $n\geq 1$ let $\mu(n)$ the Möbius function, and let $\psi(z)$ the Digamma function, see its definition and how is denoted and typeset in Wolfram Language, if you need it from this MathWorld.

While I was doing calculations I found this problem:

Calculate a good approximation of $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right).$$

My believe that why it is interesting is that Wolfram Alpha online calculator provide me approximations around $\frac{1}{2}$, with codes like this

sum mu(n)/n PolyGamma[0, 1+1/n], from n=1 to 5000

Question. My belief is that defining $$S:=\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right),$$ then $$S=\frac{1}{2}.$$ Can you provide us and justify a good approximation of $S$, or well do you know how discard that $S=\frac{1}{2}$? If do you know that it was in the literature, refers it. Many thanks.


Since $\sum_{n=1}^\infty\frac{\mu(n)}{n}=0$, we have $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\psi\left(1+\frac{x}{n}\right)=\sum_{n=1}^\infty\frac{\mu(n)}{n}\left[\psi\left(1+\frac{x}{n}\right)+\gamma\right],$$ and the series is uniformly convergent on $[0,1]$. Now $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\left[\psi\left(1+\frac{x}{n}\right)+\gamma\right]=-\sum_{n=1}^\infty\frac{\mu(n)}{n}\sum^\infty_{k=1}\zeta(k+1)\frac{(-x)^k}{n^k},$$ from the Taylor series of the Digamma function. For $|x|<1$, this is absolutely convergent, and we can change the order of summation, so we get $$-\sum^\infty_{k=1}\zeta(k+1)(-x)^k\sum_{n=1}^\infty\frac{\mu(n)}{n^{k+1}}=-\sum^\infty_{k=1}(-x)^k=\frac{x}{1+x},$$ since $$\sum_{n=1}^\infty\frac{\mu(n)}{n^{k+1}}=\frac1{\zeta(k+1)}.$$ The limit for $x\rightarrow1$ is $1/2,$ indeed.

  • $\begingroup$ Many thanks for you answer, I am going to study it. $\endgroup$
    – user243301
    Jul 21 '17 at 8:41
  • $\begingroup$ I've corrected the typo. $\endgroup$
    – user436658
    Jul 21 '17 at 8:50
  • $\begingroup$ Perfect, many thanks. $\endgroup$
    – user243301
    Jul 21 '17 at 8:51
  • $\begingroup$ This is my original problem. Equate $e^x=e^x$, where LHS is $(16)$ from this MathWorld, and RHS is the infinite product due to Sondow and Guillera, Problem 11381 The American Mathematical MONTLHY (2008). Professor Sondow share it from his homepage. Then take logarithms in our first equation, and also $\int_0^1$. Thus you've proved that our resulting RHS is also $\frac{1}{2}$. $\endgroup$
    – user243301
    Jul 21 '17 at 9:05
  • 1
    $\begingroup$ A consequence of the PNT is if $f(z) = \frac{C}{z}+\sum_{n=1}^\infty (g(nz)-\frac{1}{n})$ converges uniformly for $|z| \ge |z_0|$ then $g(z) = \sum_{n=1}^\infty \mu(n) f(nz)$ for $|z| \ge |z_0|$. Here $z_0 = 1$ and $f(z) = \frac{1}{z}(\psi(1+\frac{1}{z})+\gamma) = \frac{\gamma}{z}+\sum_{n=1}^\infty (g(nz)-\frac{1}{n})$, $g(z) = \frac{1}{1+z}$. $\endgroup$
    – reuns
    Jul 21 '17 at 19:40

This is not an answer but it is too long for a comment.

To confirm the result, I computed $$S_k=\sum_{n=1}^{10^k}\frac{\mu(n)}{n}\psi\left(1+\frac{1}{n}\right)$$ and obtained the following decimal representations $$\left( \begin{array}{cc} k & S_k \\ 1 & 0.4606256866 \\ 2 & 0.4825538785 \\ 3 & 0.4974613056 \\ 4 & 0.5012018305 \\ 5 & 0.5002812270 \\ 6 & 0.4998842081 \end{array} \right)$$ I gave up for $k=7$ (too long).

  • $\begingroup$ Many thanks, since I believe that how converges the series is a difficult issue, I don't know if your calculation is evidence about my claim, that $S=\frac{1}{2}$, or well it tell us that $S\neq\frac{1}{2}$. Any case many thanks for your contribution and help. $\endgroup$
    – user243301
    Jul 21 '17 at 8:21
  • 1
    $\begingroup$ @user243301. You are welcome ! When you plot the partial sums, it is clear that they oscillate less and less. Now, how to prove that the limit is (or not) $\frac 12$, I do not know. I hope and wish that you will get relevant answers. Cheers. $\endgroup$ Jul 21 '17 at 8:49
  • $\begingroup$ This morning I've forgotten my first equation, and now I see that that it could be much more easily deduced the result from the first equation. Any case your calculations. $\endgroup$
    – user243301
    Jul 21 '17 at 9:16

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