# 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.

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+2j+1} = \int_0^1 x^{2j} \sum_{k=1}^\infty (-1)^{k+1} x^kdx = -\int_0^1 x^{2j} \sum_{k=1}^\infty (-x)^k dx$$ 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.

• Please add some $dx$'s. – Gary May 15 at 6:49
• Just added! @Gary – epsilon-emperor May 15 at 6:51
• An alternative way is to use dlmf.nist.gov/5.7.E7 and obtain an equivalent formula in terms of the digamma function. – Gary May 15 at 6:52
• Gary, Thanks for pointing that out! @Dk65 Ah, good luck with that! – epsilon-emperor May 15 at 6:53
• @Dk65 Since $$\sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }}{{k + 2j + 1}}} \sim \frac{1}{{4j}}$$ for large $j$, you will get a divergent sum unless you sum up a finite number of terms with respect to $j$. – Gary May 15 at 7:36

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, as was done in my post here, 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*}

• You can arrive at the same result by integrating the simple identity $$- \frac{{x^{2j + 1} }}{{1 + x}} = \frac{1}{{1 + x}} - \sum\limits_{k = 0}^{2j} {( - 1)^k x^k } .$$ – Gary May 15 at 7:19
• @Gary the identity you mentioned gives such a direct solution too! Integrating the $\frac{1}{1+x}$ term yields the $\ln(2)$, while the sum gives you everything else. That said, I originally discovered how to evaluate integrals like $\int_{0}^{1}\frac{x^{2j+1}}{1+x}dx$ using $\tan$ and its reduction formula, so now I'm loyal to the $\tan$ :) – Alann Rosas May 15 at 7:27

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.

• Nice answer.It's funny to see your profile picture without the mask anymore :) – A-Level Student Jun 3 at 7:48