Lerch transcendent and harmonic number simplification I am trying to find a simplification of the following, preferably to a hypergeometric function. I have the result in Mathematica notation:
HarmonicNumber[1 + k] + (-1)^k LerchPhi[-1, 1, 2 + k] + Log[2]

which I think means the following (not sure about the Lerch transcendent):
$$H_{k+1} + (-1)^{k} \Phi (-1, 1, k+2) + \ln(2)$$
I feel like it should be possible to simplify, because evaluating LerchPhi[-1, 1, 2 + k] for different values of k yields results that contain a $(-1)^{k} \ln(2)$ term.
The reason why I think it should be in terms of a hypergeometric functions, is that I managed to derive analogous results to this one in terms of the pFq fn.
Edit: $k \geq 0$ here.
Any ideas?
 A: There is a subtle issue with Wolfram's LerchPhi, but for $\Re(2+k) > 0$ it is the same as the standard definition, see (1). So if $\Re(2+k) > 0 $ you get from (2)
$$\Phi(-1, 1, k+2) = \frac{1}{k+2}\, {_2}F_1(1,k+2, k+3, -1).$$
According to Maple the RHS can be written  in terms of the digamma function
$$\Phi(-1, 1, k+2) =\frac{1}{2}\Psi\left(\frac{3}{2}+\frac{1}{2}k\right)-\frac{1}{2}\Psi\left(1+ \frac{1}{2}k\right),$$
which may be used together with  $H_{k+1} = \Psi(k+2) + \gamma.$
Or you can use a linear transformation  15.3.4 from Abramowitz/Stegun 
$$\frac{1}{k+2}\, {_2}F_1(1,k+2, k+3, -1) =\frac{1}{2(k+2)}\, {_2}F_1(1,1, k+3, \tfrac{1}{2})$$
A: I found a simplification by looking at the numbers.
Let
$$a(k)=H_{k+1}+(-1)^k \phi (-1,1,k+2)+\log (2)$$
Then the first 10 terms (starting from k=0) are
$$\left\{2,2,\frac{8}{3},\frac{8}{3},\frac{46}{15},\frac{46}{15},\frac{352}{105},\frac{352}{105},\frac{1126}{315},\frac{1126}{315}\right\}$$
The sequence is "repetitive", i.e.
$$a(2k) = a(2k+1), k=0, 1, 2, ...$$
Furthermore, from the values we deduce (with the help of OEIS or otherwise) that
$$a(2k) = 2\sum _{i=1}^{2k} \frac{1}{2 i-1}$$
which is the harmonic partial sum of odd terms.
Hence for the (nontrivial) even terms the simplification is simply given in terms of Harmonic numbers
$$a(2k) = 2 H_{2 k+1}-H_k$$
