Find the sum of each of the following infinite series :

This geometric series is the sum of two geometric series:

$\frac{2}{5}+\frac{2}{5^{3}}+\frac{2}{5^{5}}+\cdots \infty \& \frac{3}{5^{2}}+\frac{3}{5^{4}}+\frac{4}{5^{6}}+\cdots \infty$

Sum of geometric series: $\frac{2}{5}+\frac{2}{5^{3}}+\frac{2}{5^{5}}+\cdots \infty$

Here, $a=\frac{2}{5}$

$r=\frac{\frac{2}{5^{3}}}{\frac{2}{5}}=\frac{1}{5^{2}}=\frac{1}{25}$

$\therefore \operatorname{Sum}=\frac{a}{1-r}=\frac{\frac{2}{5}}{1-\frac{1}{25}}=\frac{\frac{2}{5}}{\frac{25-1}{25}}=\frac{2 \times 25}{24 \times 5}=\frac{5}{12}$

Sum of geometric series: $\frac{3}{5^{2}}+\frac{3}{5^{4}}+\frac{4}{5^{6}}+\cdots \infty$

Here, $a=\frac{3}{5^{2}}$

$r=\frac{\frac{3}{5^{4}}}{\frac{3}{5^{2}}}=\frac{1}{5^{2}}=\frac{1}{25}$

$\therefore \operatorname{Sum}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{3}{5^{2}}}{1-\frac{1}{25}}=\frac{\frac{3}{5^{2}}}{\frac{25-1}{25}}=\frac{3 \times 25}{25 \times 24}=\frac{1}{8}$

∴Sum of the given infinite series=sum of both the series $=\frac{5}{12}+\frac{1}{8}=\frac{(5 \times 2)+(1 \times 3)}{24}$

$=\frac{10+3}{24}=\frac{13}{24}$

Sum $=\frac{13}{24}$