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.
 A: Hint:
$$\arctan(x)=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1}$$ if $|x|< 1.$  Now consider the following shift of the summation index $k\rightarrow k-1.$

 \begin{align} \arctan(x)&=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{2k-1}}{2k-1}. \space \text{Now let  $x=\frac{1}{3}$.} \end{align}


 \begin{align} \text{Your final answer should be $6\times\frac{1}{3}\arctan(\frac{1}{3})=2\arctan(\frac{1}{3})\approx 0.64350110879$} \end{align}

A: hint
$$\sum_{k=1}^n(-1)^{k-1}\frac{(\frac 19)^k}{2k-1}=$$
$$\frac 13\sum_{k=0}^{n-1}(-1)^k\frac{(\frac 13)^{2k+1}}{2k+1}$$
A: Hint:
Rewrite the finite sum as $\;\sum_{k=1}^n (-1)^{k-1}\frac{\bigl(\tfrac{1}{3}\bigr)^{\!\scriptstyle2k}}{2k-1}$, in the form
$$\sum_{k=1}^n (-1)^{k-1}\frac{x^{2k}}{2k-1}=x\sum_{k=1}^n (-1)^{k-1}\frac{x^{2k-1}}{2k-1}$$
and observe that $$\sum_{k=1}^n (-1)^{k-1}\frac{x^{2k-1}}{2k-1}=\int\sum_{k=1}^n (-1)^{k-1}x^{2(k-1)} \mathrm dx=\int\sum_{k=0}^{n-1} (-1)^{k}x^{2k} \mathrm dx$$
Now, we have the identity
$$\frac 1{1+x^2}=\sum_{k=0}^{n-1} (-1)^{k}x^{2k}+(-1)^n \frac{x^{2n}}{1+x^2}.$$
Can you proceed?
