I am sorry to say G, I do not know what it evaluates to.

But, maybe your idea will work in some respect.

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## Series Contest (SC4)

I am sorry to say G, I do not know what it evaluates to.

But, maybe your idea will work in some respect.

I wasn't saying that the other sum is in any way related. I was just sharing something I recently came across.

One thing I did notice is the series attached to the Euler sum is the alternating odd harmonic numbers. Sometimes they're called 'skew harmonic numbers' and are designated by the little minus sign on the top of the H.

So, $$1-\frac{1}{3}+\frac{1}{5}-\cdot\cdot +\frac{(-1)^{n}}{2n+1}=\sum_{k=0}^{n}\frac{(-1)^{k}}{2k+1}=H_{2n+1}^{-}$$

and $$H_{2n+1}^{-}=H_{2n+1}-H_{n}$$

so perhaps we can write the sum in question as:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}H_{2n+1}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}^{2}}{n}\tag{1}$$

Note that $H_{2n+1}=H_{2n}+\frac{1}{2n+1}$, so we can write the left one as:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}(H_{2n}+\frac{1}{2n+1})}{n}$$

for the right quadratic sum in (1), there is a generating function we can derive:

$$\sum_{n=1}^{\infty}\frac{(H_{n})^{2}}{n}x^{n}=-1/3\log(1-x)+Li_{3}(x)-Li_{2}(x)\log(1-x)$$

then let $x\to -x$ to obtain the result by letting x=1.

also, $$\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{n(2n+1)}=2\log^{2}(2)$$

a rather banal result for such an Euler sum.

The left one in (1) involving $H_{n}H_{2n}$ is the more complicated one, or so it would appear. But maybe we can derive a generating function for it as well.

Here is another Euler sum.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi'(2n+1)}{n}$$

I tried $$\frac{\pi csc(\pi z)\psi'(-2z)}{z}$$. I got a lot of $\zeta(3)$ factors and a log term.

The residue at 0 is $4\zeta(3)$

then we have the residues at the positive half-integers, the positive integers, and a simple pole at the negative integers.

Note the identity $\psi'(2n)=\psi'(2n+1)+\frac{1}{4n^{2}}$. Thus:

at the negative integers the series is $$\frac{(-1)^{n}(\psi'(2n+1)+\frac{1}{4n^{2}})}{z+n}+......$$

at the positive integers: $$\frac{(-1)^{n}}{4(z-n)^{3}}+\frac{(-1)^{n+1}\psi'(2n+1)}{z-n}+\frac{3(-1)^{n}\pi^{2}}{8(z-n)}+\cdot\cdot\cdot $$

at the positive half integers: $$\frac{\pi}{4}\cdot \frac{(-1)^{n}}{(z-\frac{2n-1}{2})^{2}}+...$$

edit:

My main issue when doing these is forgetting a negative here or there, and forgetting to divide by the (n-1)! when differentiating. That was my problem all along. But, that's all it takes.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi'(2n+1)}{n}=-\frac{\pi}{2} G+2\zeta(3)-\frac{3\pi^{2}}{16}\log(2)$$

Here is a series to consider.

find the sum:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}=\gamma \log(2)-\frac{\log^{2}(2)}{2}$$

hence show $$\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx=-1/2\log^{2}(2)$$

$$\begin{align} \sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n} &= \eta'(1) \\ &= \lim_{s \to 1} \Big( \log(2) 2^{1-s} \zeta(s) + (1-2^{1-s}) \zeta'(s) \Big) \\ &= \lim_{s \to 1} \Bigg( \log(2) 2^{1-s} \Big( \frac{1}{s-1} + \gamma + \mathcal{O}(s) \Big) + (1-2^{1-s}) \Big( -\frac{1}{(s-1)^{2}} + \mathcal{O}(1) \Big) \Bigg) \\ &= \gamma \log(2) + \lim_{s \to 1} \frac{\log(2) 2^{1-s} (s-1) - (1-2^{1-s})}{(s-1)^{2}} \\ &= \gamma \log(2) + \lim_{s \to 1} \frac{- \log^{2}(2)2^{s-1}(s-1)+\log(2)2^{s-1} - \log(2) 2^{s-1}}{2(s-1)} \\ &= \gamma \log(2) - \frac{\log^{2}(2)}{2} \end{align}$$

And since $ \displaystyle \int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \ dx = \Gamma(s) \eta(s)$,

$$ \begin{align} \int_{0}^{\infty} \frac{\log x}{1+e^{x}} \ dx &= \Gamma'(1) \eta(1) + \Gamma(1) \eta'(1) \\ &= - \gamma \log(2) + \gamma \log(2) - \frac{\log^{2}(2)}{2} \\ &= - \frac{\log^{2}(2)}{2} \end{align}$$

Problem 22?

Using the integral representation $$ \psi_{1}(z) = - \int_{0}^{1} \frac{x^{z-1} \ln x}{1-x} \ dx$$

show that $$ \sum_{n=1}^{\infty} \left(\psi_{1}(n) \right)^{2} = 3 \zeta(3)$$ where $\psi_{1}(n)$ is the trigamma function.

I thought the evaluation would turn out to be nasty, but it's not so bad.

And just to note, the sum converges since $\psi_{1}(z) \sim \frac{1}{n} $ as $ n \to \infty$.

Hey G.

I think I have the idea. Maybe.

Using the trigamma integral identity you have there, we can square it by using a double integral.

$$(\psi(n))^{2}=\int_{0}^{1}\int_{0}^{1}\frac{x^{n-1}y^{n-1}\ln(x)\ln(y)}{(1-x)(1-y)}dxdy$$

Now, take the sum of both sides. That $x^{n-1}$ and $y^{n-1}$ in the numerator of the last line are a geometric series forms for $(xy)^{n-1}$. So, we have $\displaystyle \sum_{n=1}^{\infty}(xy)^{n-1}=\frac{1}{1-xy}$

$$\sum_{n=1}^{\infty}(\psi'(n))^{2}=\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)\ln(y)}{(1-xy)(1-x)(1-y)}dxdy$$

expand these out, integrate w.r.t y. This gives:

$$\int_{0}^{1}\frac{\log(x)Li_{2}(x)}{(x-1)^{2}}dx-\pi^{2}/6\int_{0}^{1}\frac{\log(x)}{(x-1)^{2}}dx$$

I got it in the form $$ \int_{0}^{1} \frac{\ln y}{(1-y)^2} \Big(\text{Li}_{2}(y) - \zeta(2) \Big) \ dy$$ and then integrated by parts.

Yes, G that's what I got too. After I fixed my error. Thanks.

It works out the only term that contributes is $$ 3Li_{3}(1-x)|{0}^{1}=3\zeta(3)$$

Note also that $$2\int_{0}^{1}Li_{2}\left(\frac{x-1}{x}\right)\cdot \frac{1}{1-x}dx=3\zeta(3)$$

**Moderators:** Random Variable, sos440

But, maybe your idea will work in some respect.

So, $$1-\frac{1}{3}+\frac{1}{5}-\cdot\cdot +\frac{(-1)^{n}}{2n+1}=\sum_{k=0}^{n}\frac{(-1)^{k}}{2k+1}=H_{2n+1}^{-}$$

and $$H_{2n+1}^{-}=H_{2n+1}-H_{n}$$

so perhaps we can write the sum in question as:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}H_{2n+1}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}^{2}}{n}\tag{1}$$

Note that $H_{2n+1}=H_{2n}+\frac{1}{2n+1}$, so we can write the left one as:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}H_{n}(H_{2n}+\frac{1}{2n+1})}{n}$$

for the right quadratic sum in (1), there is a generating function we can derive:

$$\sum_{n=1}^{\infty}\frac{(H_{n})^{2}}{n}x^{n}=-1/3\log(1-x)+Li_{3}(x)-Li_{2}(x)\log(1-x)$$

then let $x\to -x$ to obtain the result by letting x=1.

also, $$\sum_{n=1}^{\infty}\frac{(-1)^{n}H_{n}}{n(2n+1)}=2\log^{2}(2)$$

a rather banal result for such an Euler sum.

The left one in (1) involving $H_{n}H_{2n}$ is the more complicated one, or so it would appear. But maybe we can derive a generating function for it as well.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi'(2n+1)}{n}$$

I tried $$\frac{\pi csc(\pi z)\psi'(-2z)}{z}$$. I got a lot of $\zeta(3)$ factors and a log term.

The residue at 0 is $4\zeta(3)$

then we have the residues at the positive half-integers, the positive integers, and a simple pole at the negative integers.

Note the identity $\psi'(2n)=\psi'(2n+1)+\frac{1}{4n^{2}}$. Thus:

at the negative integers the series is $$\frac{(-1)^{n}(\psi'(2n+1)+\frac{1}{4n^{2}})}{z+n}+......$$

at the positive integers: $$\frac{(-1)^{n}}{4(z-n)^{3}}+\frac{(-1)^{n+1}\psi'(2n+1)}{z-n}+\frac{3(-1)^{n}\pi^{2}}{8(z-n)}+\cdot\cdot\cdot $$

at the positive half integers: $$\frac{\pi}{4}\cdot \frac{(-1)^{n}}{(z-\frac{2n-1}{2})^{2}}+...$$

edit:

My main issue when doing these is forgetting a negative here or there, and forgetting to divide by the (n-1)! when differentiating. That was my problem all along. But, that's all it takes.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi'(2n+1)}{n}=-\frac{\pi}{2} G+2\zeta(3)-\frac{3\pi^{2}}{16}\log(2)$$

find the sum:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(n)}{n}=\gamma \log(2)-\frac{\log^{2}(2)}{2}$$

hence show $$\int_{0}^{\infty}\frac{\log(x)}{e^{x}+1}dx=-1/2\log^{2}(2)$$

And since $ \displaystyle \int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \ dx = \Gamma(s) \eta(s)$,

$$ \begin{align} \int_{0}^{\infty} \frac{\log x}{1+e^{x}} \ dx &= \Gamma'(1) \eta(1) + \Gamma(1) \eta'(1) \\ &= - \gamma \log(2) + \gamma \log(2) - \frac{\log^{2}(2)}{2} \\ &= - \frac{\log^{2}(2)}{2} \end{align}$$

Using the integral representation $$ \psi_{1}(z) = - \int_{0}^{1} \frac{x^{z-1} \ln x}{1-x} \ dx$$

show that $$ \sum_{n=1}^{\infty} \left(\psi_{1}(n) \right)^{2} = 3 \zeta(3)$$ where $\psi_{1}(n)$ is the trigamma function.

I thought the evaluation would turn out to be nasty, but it's not so bad.

And just to note, the sum converges since $\psi_{1}(z) \sim \frac{1}{n} $ as $ n \to \infty$.

I think I have the idea. Maybe.

Using the trigamma integral identity you have there, we can square it by using a double integral.

$$(\psi(n))^{2}=\int_{0}^{1}\int_{0}^{1}\frac{x^{n-1}y^{n-1}\ln(x)\ln(y)}{(1-x)(1-y)}dxdy$$

Now, take the sum of both sides. That $x^{n-1}$ and $y^{n-1}$ in the numerator of the last line are a geometric series forms for $(xy)^{n-1}$. So, we have $\displaystyle \sum_{n=1}^{\infty}(xy)^{n-1}=\frac{1}{1-xy}$

$$\sum_{n=1}^{\infty}(\psi'(n))^{2}=\int_{0}^{1}\int_{0}^{1}\frac{\ln(x)\ln(y)}{(1-xy)(1-x)(1-y)}dxdy$$

expand these out, integrate w.r.t y. This gives:

$$\int_{0}^{1}\frac{\log(x)Li_{2}(x)}{(x-1)^{2}}dx-\pi^{2}/6\int_{0}^{1}\frac{\log(x)}{(x-1)^{2}}dx$$

It works out the only term that contributes is $$ 3Li_{3}(1-x)|{0}^{1}=3\zeta(3)$$

Note also that $$2\int_{0}^{1}Li_{2}\left(\frac{x-1}{x}\right)\cdot \frac{1}{1-x}dx=3\zeta(3)$$