91Ó°ÊÓ

Using the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n→\mathrm{∞}}{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n→\mathrm{∞}}{\lim }\left|\frac{\frac{2^{−n}}{n−1}}{\frac{2^{−(n+1)}}{(n+1)+1}}\right|\\ =\underset{n→\mathrm{∞}}{\lim }|\frac{2^{−n}}{2^{−(n+1)}}×\frac{n+2}{n+1}|\\ =\underset{n→\mathrm{∞}}{\lim }|\frac{2×2^{−(n+1)}}{2^{(n+1)}}×\frac{n+2}{n+1}|\\ =\underset{n→\mathrm{∞}}{\lim }2|\frac{n+1}{n+1}+\frac{1}{n+1}|\\ =\underset{n→\mathrm{∞}}{\lim }2|1+\frac{1}{n+1}|\\ =2\end{array}$

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

.

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 $−1$,

$\begin{array}{c}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{2^n}{n+1}{\left(-1-1\right)}^n=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{2^{−n}}{n+1}\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{{(−1)}^n}{n+1}\end{array}$

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,

$\begin{array}{l}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{2^{−n}}{n+1}=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{2^{−n}}{n+1}\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{1}{n+1}\end{array}$

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$x∈[−1,3)$.