91Ó°ÊÓ

The limit of the sequence ${\mathit{ÒÏ}}_{\mathbf{n}}$as $\mathit{n}\mathbf{→}\mathbf{∞}$ is defined by the equations:

${\mathit{ÒÏ}}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$

$\mathit{ÒÏ}\mathbf{=}\underset{\mathbf{n}\mathbf{→}\mathbf{∞}}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{ÒÏ}}_{\mathbf{n}}$

If role="math" localid="1657472530818" $\mathit{ÒÏ}\mathbf{<}\mathbf{1}$, the series converges.

If $\mathit{ÒÏ}\mathbf{=}\mathbf{1}$, use a different test.

If $\mathit{ÒÏ}\mathbf{>}\mathbf{1}$, the series diverges.

The given series is $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n(x{)}^{\frac{1}{2}}}{n\ln n}$.

To Convert the series, take $y=\sqrt{x}$and simplify as follows:

$\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n(x{)}^{\frac{1}{2}}}{n\ln n}=\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n(y{)}^n}{n\ln n}$

Let $a_n=\frac{(-1{)}^n(y{)}^n}{n\ln n}$

Then, role="math" localid="1657518624392" $a_{n+1}=\frac{(-1{)}^n(y{)}^{n+1}}{(n+1)\ln (n+1)}$

Apply ratio test on the series and solve as follows:

$\underset{n→∞}{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n→∞}{\lim }\left|\frac{\frac{(-1{)}^n(y{)}^{n+1}}{(n+1)\ln (n+1)}}{\frac{(-1{)}^n(y{)}^n}{n\ln n}}\right|$

$=\underset{n→∞}{\lim }\left|\frac{(y{)}^{n+1}}{(n+1)\ln (n+1)}×\frac{n\ln n}{(y{)}^n}\right|$

$=\left|y\right|\underset{n→∞}{\lim }\left|\frac{\ln \left(n\right)+1}{\ln (n+1)+1}\right|$

$=\left|y\right|\underset{n→∞}{\lim }\left|\frac{1/n}{1/(n+1)}\right|$

Solve the equation further as follows:

$\mathrm{ÒÏ}=\left|y\right|\left(1\right)$

$\mathrm{ÒÏ}=\left|y\right|$

Then, the series converges for $\mathrm{ÒÏ}<1$.

$\left|y\right|<1$

$⇒\left|y\right|<1$

$⇒\left|\sqrt{x}\right|<1andxâ‰\yen 0$

$⇒0<x<1$

For $x=0$, the series converges to zero.

For $x=1$, the series becomes $\underset{n=0}{\overset{∞}{∑}}\frac{(-1{)}^n}{n\ln n}$and it converges by alternating series test.

Thus, the interval of convergence is $0<x<1$.