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I met an interesting function defined by series of functions, which is given as follows \begin{equation*} f(n,s)=\frac{s}{n(n+1)}-\sum_{k=0}^{+\infty}\frac{2s}{(sk+n)(sk+n+1)(sk+n+2)}. \end{equation*} Here, $n=1,2,\cdots$, and $s>0$. When I studied the difference of digamma functions, the function $f(n,s)$ was derived through the series representation of the digamma function.

It is not difficult to show $f(n,1)=0$, no matter what $n$ is. Through numerical experiments in Matlab, I guess for any given $n$, the function is strictly increasing with respect to $s$. However, I failed to give a rigorous proof. Can someone give me a hand? Thanks a lot. The Matlab code and numerical result are given below.

function y = frctseries(s,n,m) % m:number of terms in the partial sum; y = 0; for k=0:m y = y + 2*s./((s*k+n).*(s*k+n+1).*(s*k+n+2)); end y = s./(n.*(n+1)) - y; end % z1 = frctseries(s,n(1),5000); z2 = frctseries(s,n(2),5000); z3 = frctseries(s,n(3),5000); figure plot(s,z1,'g','LineWidth',1); grid on; hold on; plot(s,z2,'r','LineWidth',1); hold on; plot(s,z3,'k','LineWidth',1); hold on; L = legend('$n=1$','$n=2$','$n=5$','FontSize',14); set(L,'Interpreter','latex');

Number of terms in the partial sum is 5000.

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Too long for a comment.

At least, if $s$ is large, the series expansion of the derivative with respect to $s$ is $$\frac{1}{n (n+1)}+\frac{4 \zeta(3)}{s^3}+O\left(\frac{1}{s^4}\right)$$ When $s<n$, it is $$\frac{1}{n (n+2)}-\frac{3 n^2+6 n+2}{3 n^2 (n+1)^2 (n+2)^2}\,s+O(s^2)$$

For $s=1$, it is "easy" to show that it is positive too.

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  • $\begingroup$ Thanks for your reply! But I'm sorry. I can't complete the steps you omitted. Especially the emergence of the Riemann zeta function scares me. It seems that you divided the proof into two parts, $s<n$ and $s\geqslant n$. Did you mean $s\geqslant n$ by 'if s is large'? I think you mean s>>n. Hence I wonder if you cover all situations or if this is a rogorous proof. $\endgroup$ Commented Sep 25 at 13:22
  • $\begingroup$ @W.J. When you compute the series appears $\psi ^{(2)}(1)$ which is $=-2 \zeta (3)$. For large $s$, we do not need to care about $n$ ($s$ is large, that's all). When I tried to work with small $s$, the series expansion is problematic if the condition $s<n$ is not introduced. $\endgroup$ Commented Sep 25 at 13:31
  • $\begingroup$ In fact, the function $f(n,s)$ is transformed from $g(n,s) = \psi\left(\frac{n+2}{s}\right)+\psi\left(\frac{n}{s}\right)-2\psi\left(\frac{n+1}{s}\right)+\frac{s}{n(n+1)}$. I don't know whether $g(n,s)$ is easier for you to handle. Or you can take a look at the question which is equivalent to this one 'math.stackexchange.com/questions/5097923/…'. I will really appreciate it if you can give more information. I think you are far more capable than me. $\endgroup$ Commented Sep 25 at 13:32

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