How can I show that $$ \sum_{n=1}^\infty \frac{2}{2n+1} (\frac{j}{v})^{2n+1} < (\frac{j}{v})^3$$
given that $1\le j < v/3$. The author suggests to use a geometric series with ratio $1/3$, but I am not sure how this helps. I would appreciate if you give some help.
Construct a geometric series with $a_1= \frac {2}{3} \times (\frac{j}{v})^3$ and $q=\frac{1}{3}$. Then $\frac{2}{2n+1} (\frac{j}{v})^{2n+1} =\frac{2}{2n+1}\times (\frac{j}{v})^3 \times (\frac{j}{v})^{2(n-1)}< \frac{2}{3} \times (\frac{j}{v})^3 \times (\frac{1}{3})^{2n-2} < a_1q^{n-1} =(\frac{2}{3^n})\times (\frac{j}{v})^3$.