In problems 1-6, determine the convergence set of the given power series.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{-}\mathbf{n}}}{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}$

Using the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{2^{-n}}{n-1}}{\frac{2^{-(n+1)}}{(n+1)+1}}\right|\\ =\underset{n\to \infty }{\lim }|\frac{2^{-n}}{2^{-(n+1)}}\times \frac{n+2}{n+1}|\\ =\underset{n\to \infty }{\lim }|\frac{2\times 2^{-(n+1)}}{2^{(n+1)}}\times \frac{n+2}{n+1}|\\ =\underset{n\to \infty }{\lim }2|\frac{n+1}{n+1}+\frac{1}{n+1}|\\ =\underset{n\to \infty }{\lim }2|1+\frac{1}{n+1}|\\ =2\end{array}$

The radius of convergence is 2, therefore the range of convergence set is$|x-1|<2$

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To completely identify the convergence set, we have to check whether the boundary pointsand 3 are included in the set or not.

Checking at ,$x=-1$by substituting the value of X by $-1$,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{2^n}{n+1}{\left(-1-1\right)}^n=\sum _{n=0}^{\infty }\frac{2^{-n}}{n+1}\\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n+1}\end{array}$

The above series is an alternating harmonic series, which is convergent in nature, thus the point is included in the convergent set.

Similarly, checking at $x=3$, by substituting the value of x by 3,

$\begin{array}{l}\sum _{n=0}^{\infty }\frac{2^{-n}}{n+1}=\sum _{n=0}^{\infty }\frac{2^{-n}}{n+1}\\ =\sum _{n=0}^{\infty }\frac{1}{n+1}\end{array}$

The above series is a harmonic series, which is divergent in nature, thus the point 3 is excluded from the convergent set.

The convergent set for the given power series is$x\in [-1,3)$.