Finding the interval of convergence and its radius of the given series

Exercise :

Find the radius and the interval of convergence of $\sum\limits_{k=0}^{\infty }{\frac{{{x}^{k}}}{\left( k+1 \right){{3}^{k}}}}$

Solution : This series is convergent by ratio test when $\left| x \right|<3$

Using ratio test

\[\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{\frac{{{x}^{k+1}}}{\left( k+2 \right){{3}^{k+1}}}}{\frac{{{x}^{k}}}{\left( k+1 \right){{3}^{k}}}} \right|=\underset{k\to \infty }{\mathop{\lim }}\,\frac{{{3}^{k}}\left( k+1 \right){{x}^{k+1}}}{{{x}^{k}}\left( k+2 \right){{3}^{k+1}}}=\underset{k\to \infty }{\mathop{\lim }}\,\frac{\left( k+1 \right)x}{3\left( k+2 \right)}=\frac{x}{3}\]

So $\left| x \right|<1\Leftrightarrow \left| \frac{x}{3} \right|<1\Leftrightarrow \left| x \right|<3$

Hence the series is convergent with interval of convergence is $\left| x \right|<3$

With radius of convergence is 3

2nd method By Cauchy root Test

Let ${{u}_{k}}=\frac{{{x}^{k}}}{\left( k+1 \right){{3}^{k}}}\Leftrightarrow {{\left( {{u}_{k}} \right)}^{1/k}}={{\left( \frac{{{x}^{k}}}{\left( k+1 \right){{3}^{k}}} \right)}^{1/k}}=\frac{x}{{{\left( k+1 \right)}^{1/k}}3}$

So $\underset{k\to \infty }{\mathop{\lim }}\,{{\left( {{u}_{k}} \right)}^{1/k}}=\underset{k\to \infty }{\mathop{\lim }}\,\frac{x}{3{{\left( k+1 \right)}^{1/k}}}=\frac{x}{3}$  since ($\underset{k\to \infty }{\mathop{\lim }}\,{{\left( k+1 \right)}^{1/k}}=1$

Let $y={{\left( k+1 \right)}^{1/k}}\Leftrightarrow \ln y=\frac{1}{k}\ln \left( k+1 \right)\Leftrightarrow y={{e}^{\frac{\ln \left( k+1 \right)}{k}}}$

Since $e$ is continuous function hence $\underset{k\to \infty }{\mathop{\lim }}\,y={{e}^{\underset{k\to \infty }{\mathop{\lim }}\,\frac{\ln \left( k+1 \right)}{k}}}={{e}^{0}}=1$ )

So $\left| x \right|<1\Leftrightarrow \left| \frac{x}{3} \right|<1\Leftrightarrow \left| x \right|<3$ with radius of convergence is 3