Not quite a telescoping sum (1)

Let us first logarithmically differentiate the Weierstrass Product : $\cos(\pi x) = \prod_{n=0}^\infty \left(1 - \frac{4x^2}{(2n+1)^2} \right)$ to get $\sum_{n=0}^\infty \frac1{ (2n+1)^2 - (2x)^2} = \frac\pi{8x} \tan(\pi x)$ and noticing that $\frac1{8n+1} - \frac1{8n+7} = \frac38 \cdot \frac1{(2n+1)^2 - \left(2\cdot \frac38 \right)^2 }$

we can conclude that $\sum_{n=0}^\infty \left ( \frac1{8n+1} - \frac1{8n+7} \right) = \frac38 \sum_{n=0}^\infty \frac1{(2n+1)^2 - \left(\frac34 \right)^2 } = \frac38 \cdot \frac\pi{8 \cdot \frac38} \tan\left( \pi \cdot \frac38 \right) = \frac{\pi}{8}(1 + \sqrt2)$

For $\lvert x \rvert <1$ consider the following geometric sequences:

$\frac{1}{1-x^8} = \lim_{n \to \infty} \sum_{k=0}^{n} x^{8k} \ \dots \ (1)$ $\frac{x^6}{1-x^8} = \lim_{n \to \infty} \sum_{k=0}^{n} x^{8k+6} \ \dots \ (2)$

Integrating both sides with respect to $x$ from 0 to 1 gives:

$\lim_{t \to 1^{-}} \int_{0}^{t} \frac{1}{1-x^8} dx= \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{8k+1} \ \dots \ (3)$ $\lim_{t \to 1^{-}} \int_{0}^{t} \frac{x^6 }{1-x^8}dx = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{1}{8k+7} \ \dots \ (4)$

Subtracting (3) and (4) gives:

$\lim_{n \to \infty} \sum_{k=0}^{n} \left[\frac{1}{8k+1}-\frac{1}{8k+7}\right]= \lim_{t \to 1^{-}} \int_{0}^{t} \left(\frac{1-x^6}{1-x^8}\right) dx=\boxed{ \frac{\pi \left(\sqrt{2}+1\right)}{8}}$

The steps of evaluation of the right-hand side limit are left out. I will add those details if requested.

We can rewrite the above sum as (1-1/7)+sigma from n= 1 to infinity (1/8n-1/(8n+7))-sigma from n=1 to infinity (1/8n-1/(8n+1)) ,then recognizing that there are digamma functions hiding in both the sum we have (8/7+ digamma(1-1/8)-digamma (1/8)-8) all divided by 8 . Simplifying by using the reflection equation 1/7-1+πcot (π/8) ,since cot(π/8)=2^1/2+1. Hence we have by inputting it to the original sum, 1-1/7+1/7-1+(π(2^1/2+1)/8)=(π(2^1/2+1))/8,a=2,b=1 and c=8.Adding it gives 11 as our result.(sorry for the messy calculation.)