Asymptotic evaluation of a series

Question

I am reading a paper in which the authors derive the asymptotics for the following series: $$ \sum_{k=0;\,k+=2}^{N(1+m)-2}\left(1-\frac{k}{2}\log{\frac{k+2}{k}}\right)\sim \frac{1}{2}\log{\left(\pi N(m+1)\right)+\mathcal{O}(1/N)}\,, $$ where $m$ is a real number and $-1\le m\le 1$. For completeness I report here what they write about this formula: "where the last line can be obtained by recognizing that the sum can be written as a convergent part plus a divergent sum which is the harmonic number, and then using the asymptotic expression of the latter".

I am looking for an argument to get the result they provide, do you have any suggestion?

Answer 1

Suggested method: Using the quadratic approximation $\log(1+x) = x - \frac{x^2}2 + O(x^3)$, rewrite $$ \log\frac{k+2}k = \log\biggl( 1+\frac2k \biggr) = \frac2k - \frac2{k^2} + O\biggl( \frac1{k^3} \biggr) $$ and examine the sum that results.

Answer 2

$$S_p=\sum_{k=0}^{p}\left(1-\frac{k}{2} \, \log \left(\frac{k+2}{k}\right)\right)$$ $$S_p=1+p+\log (\Gamma (p+3))+\frac{1}{2} \left(\zeta ^{(1,0)}(-1,p+1)-\zeta^{(1,0)}(-1,p+3)\right)$$

Expanding for large values of $q$ $$\zeta^{(1,0)}(-1,q)=\frac{1}{4} q^2 (2 \log (q)-1)-\frac{1}{2} q \log (q)+\frac{1}{12} (\log (q)+1)+$$ $$\frac{1}{720 q^2}-\frac{1}{5040 q^4}+O\left(\frac{1}{q^6}\right)$$

Using the above and continuing with Taylor expansion $$S_p=\log \left( p\sqrt{\frac{2 \pi}{e}}\right)+\frac{11}{6 p}-\frac{7}{4 p^2}+O\left(\frac{1}{p^3}\right)$$

Trying for $p=1000$, the result is in an absolute error of $2.29\times 10^{-9}$.