$\sum_{n=1} ^\infty \frac{8}{n15^n}$ without calculator I just can't figure out which function to use to get the sum. I tried with ln, but that gives me an alternating series.

Hint
Consider $$A=\sum_{n=1} ^\infty \frac{8}{n15^n}={8}\sum_{n=1} ^\infty \frac{1}{n15^n}={8}\sum_{n=1} ^\infty \frac{x^n}{n}$$ where $x=\frac 1 {15}$. 
You can recongnize that the summation is just Taylor expansion of $-\log(1-x)$ from which the problem becomes simple.
We then have $$A=-8\log(1-\frac 1 {15})=-8\log(\frac {14} {15})=8\log(\frac {15} {14})$$ Now, with no calculator, remember the fast convergent expansion $$\log \left(\frac{1+y}{1-y}\right)=2\left( y+\frac{ y^3}{3}+\frac{ y^5}{5}+\frac{ y^7}{7}+O\left(y^9\right)\right)$$ Setting $\frac{1+y}{1-y}=\frac {15} {14}$ gives $y=\frac {1} {29}$. So, $$8\log(\frac {15} {14})=16\left(\frac {1} {29}+\frac{1}{73167}+\frac{1}{102555745}+\frac{1}{120749134163}+\cdots\right)$$ Just using the first and second terms probably gives a sufficient approximation $$16\left(\frac {1} {29}+\frac{1}{73167}\right)=\frac{40384}{73167}\approx 0.5519428158$$ while the exact value would be $\approx 0.5519429719$.
Instead of Taylor series, you could use Padé approximants; a simple one would be $$\log(1+x)\approx \frac{x+\frac{x^2}{2}}{1+x+\frac{x^2}{6}}$$ which, for $x=\frac 1 {14}$, would lead to $$A\approx \frac{696}{1261}=0.5519429025$$
