On a relation between Euler sums

Prove that:

$$\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1} = \frac{\pi \log 2}{8} - \frac{\mathcal{G}}{2}$$

Solution

An Euler non linear sum

Let $\mathbb{N} \ni m \geq 2$. Then the following formula

$$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n^{(m)}}{n^m}=\frac{\zeta^2(m)+\zeta(2m)}{2}$$

holds.

Solution

An Euler trigonometric sum

Prove that:

$$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} \cos \frac{n \pi}{3}= -\frac{\pi^2}{36}$$

Solution

Logarithmic integral

Evaluate the integral:

$$\int_{0}^{1}\frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$

Solution

Logarithmic integral and Euler sums

Evaluate:
$$\int_0^1 \frac{\ln (1+x^2)}{1+x}\, {\rm d}x$$

Solution

Alternating Euler Sum

Evaluate the sum:

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

Solution:

Euler Sum

Let \( \mathcal{H}_n \) denote the \( n \) th harmonic number. Evaluate the sum:

$$ \mathcal{S}=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} $$

Answer: \( \displaystyle \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2}=2\zeta(3) \)

Proof: