How can I derive the first two terms of the asymptotic expansion of $f(n)=\sum_{k=1}^\infty [(-1)^k/k]\ln(n^2+k)$ at $n \to +\infty$?

Question

I am struggling with the problem in the title of this post. I have tried many different methods, but nothing has worked so far. I only managed to derive the first term of the asymptotic expansion:

$f(n)=\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln(n^2+k) =\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln(n^2) + \sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln\left(1+\frac{k}{n^2}\right) = -\ln (2) \ln(n^2)+ o(1)$, $\quad n \to +\infty$,

where I used $\ln(n^2+k)=\ln\left( n^2\left(1+\frac{k}{n^2}\right)\right)= \ln(n^2) +\ln \left(1+\frac{k}{n^2}\right)$.

How can I derive the next expansion term?

Answer 1

You found that $$f(n)=-2\ln(2)\ln (n)+\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln\left(1+\frac{k}{n^2}\right)$$ Using the Frullani integral $$\ln a=\int_0^\infty\frac{e^{-t}-e^{-at}}tdt$$ $$S(n)=\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\ln\left(1+\frac{k}{n^2}\right)=\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\int_0^\infty\frac{e^{-t}-e^{-t(1+\frac k{n^2})}}tdt$$ $$=\int_0^\infty\frac{e^{-t}}t\sum_{k=1}^\infty\frac{(-1)^k}k\left(1-e^{-\frac{tk}{n^2}}\right)dt=\int_0^\infty\frac{e^{-t}}t\ln\frac{1+e^{-\frac t{n^2}}}2dt$$ Decomposing $\ln (1+e^{-\frac t{n^2}})=\ln\left(2-\frac t{n^2}+\frac{t^2}{2n^4}+O(1/n^6)\right)$ $$S(n)=\int_0^\infty\frac{e^{-t}}t\left(-\frac t{2n^2}+\frac {t^2}{4n^4}-\frac {t^2}{8n^4}+O\left(\frac1{n^6}\right)\right)dt$$ $$=-\frac1{2n^2}+\frac1{8n^4}+O\left(\frac1{n^6}\right)$$ Therefore, $$\boxed{\,\,f(n)=-2\ln(2)\ln(n)-\frac1{2n^2}+\frac1{8n^4}+O\left(\frac1{n^6}\right)\,\,}$$

Answer 2

We can derive a complete asymptotic series starting from Svyatoslav's integral formula. Note that $$ \int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{t}\log \left( {\frac{{1 + {\rm e}^{ - t/n^2 } }}{2}} \right){\rm d}t} = \int_0^{ + \infty } {\frac{{{\rm e}^{ - n^2 s} }}{s}\log \left( {\frac{{1 + {\rm e}^{ - s} }}{2}} \right){\rm d}s} , $$ and $$ \log \left( {\frac{{1 + {\rm e}^{ - s} }}{2}} \right) = - \int_0^s {\frac{{{\rm d}t}}{{{\rm e}^t + 1}}} = - \frac{1}{2}\sum\limits_{m = 0}^\infty {\frac{{E_m (0)}}{{(m + 1)!}}s^{m + 1} } = - \frac{s}{2} + \sum\limits_{m = 1}^\infty {\frac{{(2^{2m} - 1)B_{2m} }}{{2m(2m)!}}s^{2m} } $$ for $|s|<\pi$. Here the $E_m(x)$ are the Euler polynomials and the $B_m$ are the Bernoulli numbers (cf. $(24.4.26)$ and $(24.2.2)$). Then, by Watson's lemma, $$ \int_0^{ + \infty } {\frac{{{\rm e}^{ - n^2 s} }}{s}\log \left( {\frac{{1 + {\rm e}^{ - s} }}{2}} \right){\rm d}s} \sim - \frac{1}{{2n^2 }} + \sum\limits_{m = 1}^\infty {\frac{{(2^{2m} - 1)B_{2m} }}{{(2m)^2 n^{4m} }}} $$ as $n\to+\infty$. Accordingly, $$ f(n) \sim - (2\log 2)\log n - \frac{1}{{2n^2 }} + \sum\limits_{m = 1}^\infty {\frac{{(2^{2m} - 1)B_{2m} }}{{(2m)^2 n^{4m} }}} $$ as $n\to+\infty$.

Answer 3

Just for the fun (because so good answers from @Svyatoslav and @Gary).

$$f(n)=\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}\log(n^2+k) $$ $$f'(n)=2\sum\limits_{k=1}^{\infty} (-1)^k\frac{ n}{k \left(k+n^2\right)}=\frac 2n \left(\Phi \left(-1,1,n^2+1\right)-\log (2)\right)$$ where appears the Hurwitz-Lerch transcendent function $$f(n)=-2\log (2) \log (n)+2\int \frac{\Phi \left(-1,1,n^2+1\right)}{n}\,dn$$ which cannot be integrated.

But $$ \frac 12 < m\,\Phi \left(-1,1,m\right)< 1$$ So $$\log \left(\frac{n\, 2^{-2 \log (n)}}{\sqrt{n^2+1}}\right)< f(n) < \log \left(\frac{n^2\ 2^{-2 \log (n)}}{n^2+1}\right)$$ Which seems to be decent. $$\left( \begin{array}{cccc} n & \text{left bound} & \text{summation}& \text{right bound} \\ 10^1 & -3.19703589584 & -3.19704823073 & -3.20201106127 \\ 10^2 & -6.38417145833 & -6.38417145958 & -6.38422145583 \\ 10^3 & -9.57618269125 & -9.57618269125 & -9.57618319125 \\ 10^4 & -12.7682429267 & -12.7682429267 & -12.7682429317 \\ 10^5 & -15.9603036521 & -15.9603036521 & -15.9603036522 \\ 10^6 & -19.1523643825 & -19.1523643825 & -19.1523643825 \\ \end{array} \right)$$