How to prove the asymptotic expansion $\overline{H}_n \sim \log(2) -(-1)^n\left (\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4}\mp\ldots\right)$? It is well-known that the harmonic sum $H_{n}= \sum_{k=1}^{n}\frac{ 1}{k}$ has the following asymptotic expansion for $n\to\infty$ 
$$H_n = \sum_{k=1}^{n}\frac{1}{k}\sim \gamma+\log \left(n\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120 n^4}-\frac{1}{252 n^6}\pm \ldots\tag{1}$$
The alternating harmonic sum is defined as
$$\overline {H}_{n} = \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}\tag{2}$$
and we ask for its asymptotic expansion.
At first I tried to use the representation
$$\overline{H}_{n} =\log (2)+ (-1)^{n+1} \Phi (-1,1,n+1)\tag{3}$$
where $ \Phi (z,s,a)=\sum_{k=0}^{\infty} \frac{z^k}{(k+a)^s}$ is a special function called Lerch transcendent (https://en.wikipedia.org/wiki/Lerch_zeta_function) which is just the tail of the expansion of $\log(2)$ starting at the $(n+1)$st term. But I couldn't find the asymptotics of $\Phi$. Also Mathematica wouldn't do it.
So I came up with another idea and found 
$$\overline{H}_{n} \sim \log(2) -(-1)^n \left(\frac{1}{2n}-\frac{1}{4 n^2} +\frac{1}{8n^4} - \frac{1}{4n^6}+\ldots\right)\tag{4}$$
I have looked up possibly related proofs. This reference contains two of them. 
Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$
But mine was still different. 
What would be your proof?
 A: My idea was to express $\overline{H}_k$ by $H_k$ and then use the asyptotic expansion of $H_k$.
Indeed, $\overline{H}_n$ can be expressed as follows ($m=1,2,3,\ldots$}
$$\overline{H}_{2m} = H_{2m} -H_{m}\tag{5a}$$
$$\overline{H}_{2m+1} = H_{2m+1} -H_{m}\tag{5b}$$
The (simple) proof is left as an exercise to the reader.
For the asymptotic expressions of the even version we find from $(1)$
$$\overline{H}_{2m}\overset{m\to\infty,m->\frac{n}{2}}  = \log (2) \\-\frac{1}{2 n}+\frac{1}{4 n^2}-\frac{1}{8 n^4}+\frac{1}{4 n^6} -\frac{17}{16 n^8}\pm\ldots\tag{6a}$$
For the odd version we have, to begin with,
$$\overline{H}_{2m+1}\overset{m\to\infty, m->\frac{n-1}{2}}=\log (2)
\\
+\frac{1}{2 (n-1)}-\frac{3}{4 (n-1)^2}+\frac{1}{(n-1)^3}-\frac{9}{8 (n-1)^4}+\frac{1}{(n-1)^5}-\frac{3}{4 (n-1)^6}
\\
+\frac{1}{(n-1)^7}-\frac{33}{16 (n-1)^8}+\frac{1}{(n-1)^9}\mp\ldots$$
Taking the asymptotics of this in turn we get
$$\overline{H}_{2m+1}\overset{m\to\infty, m->\frac{n-1}{2}}=\log (2)\\+ \frac{1}{2 n}-\frac{1}{4 n^2}+\frac{1}{8 n^4}-\frac{1}{4 n^6}+\frac{17}{16 n^8}\mp\ldots\tag{6b}$$
Finally, combining $(6a)$ and $(6b)$ gives the expression $(4)$ of the OP.
Combining this with $(3)$ we have also derived the asymptotics of the Lerch $\Phi$ function from that of the harmonic number.
