Alternating series summation I'm curious if the following series can be summed up
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+2j+1}$$
I tried to apply the typical alternating series summation for series of this form through the Taylor series of the $\log(1+x)$, but that $2j+1$ term really messes it up.
 A: $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+2j+1} = \int_0^1 x^{2j} \sum_{k=1}^\infty (-1)^{k+1} x^kdx = -\int_0^1 x^{2j} \sum_{k=1}^\infty  (-x)^k dx$$
Since $|x| < 1$, we have
$$\sum_{k=1}^\infty  (-x)^k = \frac{-x}{1+x}$$
The question then boils down to
$$\int_0^1 \frac{x^{2j+1}}{1+x} dx $$
From what I know,
$$\int_0^1 \frac{x^{2n+1}}{1+x}dx = \frac{1}{2} \left(H_{n+\frac12} - H_n \right)$$ where $n > -1$ and $n \in\mathbb R$. Assuming $j$ satisfies the required conditions, I would conclude
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+2j+1} =  \frac{1}{2} \left(H_{j+\frac12} - H_j \right)$$ where $H_n$ is the $n$th harmonic number.
A: We can take @epsilon-emperor's answer one step further. Making the substitution $x=\tan^2(u)$ in the integral they derived gives
\begin{align*}
\int_{0}^{1}\frac{x^{2j+1}}{1+x}dx &= \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan^2 u\right)^{2j+1}}{1+\tan^2(u)}\cdot 2\tan(u)\sec^2(u)\text{ }du\\
&= 2\int_{0}^{\frac{\pi}{4}}\frac{\tan^{2(2j+1)+1}(u)}{\sec^2(u)}\cdot\sec^2(u)\text{ }du\\
&= 2\int_{0}^{\frac{\pi}{4}}\tan^{2(2j+1)+1}(u)\text{ }du
\end{align*}
Using the reduction formula for $\int\tan^n(x)dx$ together with mathematical induction, it can be shown that for every integer $n\geq 1$,
$$\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$$
Thus, the exact sum of your series is
\begin{align*}
2\int_{0}^{\frac{\pi}{4}}\tan^{2(2j+1)+1}(u)\text{ }du &= 2\cdot(-1)^{2j+1}\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{2j+1}\frac{(-1)^{k-1}}{k}\right)\\
&= -\ln(2)+\sum_{k=1}^{2j+1}\frac{(-1)^{k-1}}{k}
\end{align*}
A: An Evaluation in Terms of Extended Harmonic Numbers
$$
\begin{align}
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k+2j+1}
&=\lim_{N\to\infty}\left(\sum_{k=1}^{2N}\frac1{k+2j+1}-\sum_{k=1}^N\frac2{2k+2j+1}\right)\tag{1a}\\
&=\lim_{N\to\infty}\sum_{k=1}^N\left(\frac1k-\frac2{2k+2j+1}\right)\\
&-\lim_{N\to\infty}\sum_{k=1}^{2N}\left(\frac1k-\frac1{k+2j+1}\right)\\
&+\lim_{N\to\infty}\sum_{k=N+1}^{2N}\frac1k\tag{1b}\\[6pt]
&=H_{j+1/2}-H_{2j+1}+\log(2)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: write as a limit and break into sums over even $k$ and sums over all $k$
$\text{(1b)}$: write as partial sums of extended Harmonic Numbers
$\phantom{\text{(1b):}}$ distribute the limit since the sums now converge
$\text{(1c)}$: evaluate the limits
Note that $H_{2j+1}=\overbrace{\quad\frac{H_j}2\quad}^\text{even terms}+\overbrace{\frac{H_{j+1/2}-H_{-1/2}}2}^\text{odd terms}$ and, as shown in $(11)$ of this answer, $H_{-1/2}=-2\log(2)$. Therefore, $\text{(1c)}$ equals $\frac12(H_{j+1/2}-H_j)$, as shown in epsilon emperor's answer

Acceleration Using Integration by Parts
In this answer, this sum is called $AHT(2j+1)$. In $(13)$ of that answer, it is shown that
$$
AHT(m)=\sum_{k=1}^\infty\frac1{k\binom{m+k}{k}2^k}\tag2
$$
Therefore,
$$
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k+2j+1}=\sum_{k=1}^\infty\frac1{k\binom{2j+1+k}{k}2^k}\tag3
$$
The sum on the right side of $(3)$ converges far faster than the sum on the left.
