Expected value. Problem with series. We have pawn on the infinite board. We roll a dice. When we roll six we can roll dice again. We move pawn as many times as we threw dots. Pawn move's What's the expected value of pawn moves in one turn ? I figure out that expected value will be equal to : $$ E(X)=\frac16 \cdot (1+2+3+4+5)+\frac{1}{36} \cdot (7+8+9+10+11) + \frac{1}{216} \cdot (13+14+15+16+17)+... $$ And i have some trouble to calculate above series.

First note that $1+2+3+4+5 = 15$. So your series becomes:

\begin{align*} \frac{1}{6}\cdot 15 + \frac{1}{6^2} \cdot (6\cdot 5+15)+ \frac{1}{6^3}(12\cdot 5+15)+\cdots &= \sum_{k=1}^\infty \frac{1}{6^k}(6\cdot(k-1)\cdot 5+15)\\\ &=30\sum_{k=1}^\infty \frac{(k-1)}{6^k} + 15\sum_{k=1}^\infty\frac{1}{6^k}\\\ &=5\sum_{k=0}^\infty \frac{k}{6^k}+\frac{15}{6}\sum_{k=0}^\infty \frac{1}{6^k}. \end{align*}

The last term is just a geometric series. The sum of $\frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=\sum_{k=0}^\infty \frac{k}{6^k}=0+\frac{1}{6}\frac{2}{6^2}+\frac{3}{6^3}+\cdots$. Then $\frac{1}{6}S=0+\frac{1}{6^2}+\frac{2}{6^3}+\frac{3}{6^4}+\cdots$, so we have $S-\frac{1}{6}S=\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+\cdots = \frac{1}{6}\sum_{k=0}^\infty \frac{1}{6^k}.$