What is the partial sum $\sum_{k=1}^n (-1)^{k-1}\frac{\left(\frac{1}{9}\right)^k}{2k-1}$? I have a problem where, after some work, I've arrived at
$$6 \times \lim_{n \to ∞} \sum_{k=1}^n (-1)^{k-1}\frac{\left(\frac{1}{9}\right)^k}{2k-1}$$
and I need to find the partial sum $$\sum_{k=1}^n (-1)^{k-1}\frac{\left(\frac{1}{9}\right)^k}{2k-1}.$$
to calculate the above limit, but I'm having trouble finding it.
I know that if this was simply a series with the $\left(\frac{1}{9}\right)^k$ term, I would just use the geometric series formula, but there's an elusive alternating term as well as the $2k-1$ term.

hint
$$\sum_{k=1}^n(-1)^{k-1}\frac{(\frac 19)^k}{2k-1}=$$
$$\frac 13\sum_{k=0}^{n-1}(-1)^k\frac{(\frac 13)^{2k+1}}{2k+1}$$
