Question

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. 

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}\mathbf{(}\mathbf{x}{\mathbf{)}}^{\frac{\mathbf{1}}{\mathbf{2}}}}{\mathbf{nlnn}}$

Short answer

The series converges when$\rho =\sqrt{x}<1$.

The interval of convergence is $0<x<1$.

Step-by-step solution

Concept of Ratio Test  

The limit of the sequence ${\mathit{\rho }}_{\mathbf{n}}$as $\mathit{n}\mathbf{\to }\mathbf{\infty }$ is defined by the equations:

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

$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}}$

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

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

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

Find the ratio an+1an.

The given series is $\sum _{n=0}^{\infty }\frac{(-1{)}^n(x{)}^{\frac{1}{2}}}{n\ln n}$.

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

$\sum _{n=0}^{\infty }\frac{(-1{)}^n(x{)}^{\frac{1}{2}}}{n\ln n}=\sum _{n=0}^{\infty }\frac{(-1{)}^n(y{)}^n}{n\ln n}$

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:

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

$=\underset{n\to \infty }{\lim }\left|\frac{(y{)}^{n+1}}{(n+1)\ln (n+1)}\times \frac{n\ln n}{(y{)}^n}\right|$

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

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

Solve the equation further as follows:

$\rho =\left|y\right|\left(1\right)$

$\rho =\left|y\right|$

Then, the series converges for $\rho <1$.

$\left|y\right|<1$

$\Rightarrow \left|y\right|<1$

$\Rightarrow \left|\sqrt{x}\right|<1andx\ge 0$

$\Rightarrow 0<x<1$

 Step 3: Check the end points.  

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

For $x=1$, the series becomes $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{n\ln n}$and it converges by alternating series test.

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