Weird Summation

Nice Solution!

The last series can be also transformed to a telescoping series:

$\displaystyle \sum_{n=0}^{\infty} \frac{n2^n}{(n+2)!} = \sum_{n=0}^{\infty} \frac{(n+2-2)2^n}{(n+2)!} = \sum_{n=0}^{\infty} \frac{2^n}{(n+1)!} - \sum_{n=0}^{\infty} \frac{2^{n+1}}{(n+2)!}$

Telescoping to 1

You need to be more careful about this derivation! You write $\sum_{j=0}^\infty \frac{1}{3j+2} \; = \; \int_0^1 \frac{x}{1-x^3}\,dx \qquad \qquad \sum_{j=0}^\infty \frac{1}{3j+4} \; = \;\int_0^1 \frac{x^3}{1-x^3}\,dx$ and subtract the two to obtain your first result.

However, both of these "formulae" state that $\infty = \infty$ , since both series and both integrals diverge; subtracting the two "formulae", as you do, is meaningless (even though the end result is the correct answer!).

Do your initial calculations with finite sums, and only take the limit later. Thus $\begin{array}{rcl} \displaystyle\sum_{j=0}^{N-1} \frac{1}{3j+2} & = &\displaystyle \int_0^1 \sum_{j=0}^{N-1} x^{3j+1}\,dx \; = \; \int_0^1 \frac{x(1 - x^{3N})}{1-x^3}\,dx \\ \displaystyle\sum_{j=0}^{N-1} \frac{1}{3j+4} & = &\displaystyle \int_0^1 \sum_{j=0}^{N-1} x^{3j+3}\,dx \; = \; \int_0^1 \frac{x^3(1-x^{3N})}{1-x^3}\,dx \end{array}$ These integrals are OK, since the factor of $1 - x^{3N}$ in the numerator cancels out the singularity at $x=1$ in the denominator in each case. We can now subtract these equations to deduce that $\sum_{j=0}^{N-1}\left(\frac{1}{3j+2} - \frac{1}{3j+4}\right) \; = \; \int_0^1 \frac{(x - x^3)(1 - x^{3N})}{1 - x^3}\,dx \; = \; \int_0^1 \frac{x(1+x)(1 - x^{3N})}{1 + x + x^2}\,dx$ This last integral contains no singularity at $x=1$ , and we can now let $N \to \infty$ to deduce that $\sum_{j=0}^{N-1}\left(\frac{1}{3j+2} - \frac{1}{3j+4}\right) \; = \; \int_0^1 \frac{x(1+x)}{1 + x + x^2}\,dx \; = \; 1 - \frac{\pi}{3\sqrt{3}} \;,$ and the rest of the result follows.