Answer
$ln(1+\frac{3}{5})$
or
$ln(8)-ln(5)$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{3^{n}}{5^{n}(n)}$
$=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{3^{n}(1/5^{n})}{n}$
$=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{(\frac{3}{5})^{n}}{n}$
As we know $ln(1+x)=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{x^{n}}{n}$
Thus,
$ln(1+\frac{3}{5})=\Sigma_{n=0}^{\infty}{(-1)^{n-1}}\dfrac{(\frac{3}{5})^{n}}{n}$
or
$ln(1+\frac{3}{5})=ln(\frac{8}{5})=ln(8)-ln(5)$
Hence,
$ln(1+\frac{3}{5})$
or
$ln(8)-ln(5)$
