91Ó°ÊÓ

The ratio test for convergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{→}\mathbf{∞}}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{<}\mathbf{1}$

The ratio test for divergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{→}\mathbf{∞}}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{>}\mathbf{1}$

${\text{n}}^{\text{th}}\sqrt{b^2-4ac}$term of the series is,$a_n=\frac{{\left(x\right)}^n}{\ln (n+1)}$

The magnitude of ${\text{n}}^{\text{th}}$terms is:

$\left|a_n\right|=\frac{{\left(x\right)}^n}{\ln (n+1)}$ …… (1)

The magnitude of $(n+1{)}^{th}$terms is calculated as follows:

$\left|a_{n+1}\right|=\frac{{\left(x\right)}^{n+1}}{\ln (n+2)}$ …… (2)

Take the ratio of equation (1) by equation (2) as follows:

$$\begin{gathered} \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{{\left(x\right)}^{n+1}}{\ln (n+2)}}{\frac{{\left(x\right)}^n}{\ln (n+1)}} \\ \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{x{\left(x\right)}^n}{\ln (n+2)}}{\frac{{\left(x\right)}^n}{\ln (n+1)}} \\ \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\ln (n+1)}{\ln (n+2)}x \end{gathered}$$

Apply the limit in the above ratio as follows:

$li{m}_{n→∞}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n→∞}\left(\frac{\ln (n+1)}{\ln (n+2)}x\right)$

$$\begin{gathered} li{m}_{n→∞}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left(\frac{\frac{1}{(n+1)}}{\frac{1}{(n+2)}}x\right) \\ li{m}_{n→∞}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n→∞}\left(\frac{(n+2)}{(n+1)}x\right) \\ li{m}_{n→∞}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n→∞}\left(\frac{\left(1+\frac{2}{-}\right)}{\left(1+\frac{1}{n}\right)}x\right) \\ li{m}_{n→∞}\frac{\left|a_{n+∞}\right|}{\left|a_n\right|}=x \end{gathered}$$

The series converges for$\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$ or $x<1$.

The series diverges for$\frac{\left|a_{n+1}\right|}{\left|a_n\right|}>1$ or $x>1$.

For, $x=1$the series becomes:

$\underset{n=1}{\overset{∞}{∑}}\frac{{\left(1\right)}^n}{\ln (n+1)}$

This series is divergent using a comparison test.

For, $x=-1$the series becomes:

$\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{\ln (n+1)}$

This is convergence alternating series.

Thus, the interval of converges is.$-1â\copyright ½x<1$