Question

What is $x_0$if the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$converges conditionally at both $x=-4$and $x=8$.

Short answer

Ans:$x_0=2$

Step-by-step solution

Step 1. Given information.

given,

$\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$

Step 2. To find the interval of convergence for the power series, let us first assume bk=akx−x0k, so bk+1=ak+1x−x0k+1

Therefore,

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}{\left(x-x_0\right)}^{k+1}}{a_k{\left(x-x_0\right)}^k}\right|\\ =\underset{k\to \mathrm{\infty }}{lim} \frac{a_{k+1}}{a_k}\left|x-x_0\right|\\ =\frac{a_{k+1}}{a_k}\underset{k\to \mathrm{\infty }}{lim} \left|x-x_0\right|\end{array}$

Step 3. Now,

Here the limit is $\left|x-x_0\right|$. So, by the ratio test of absolute convergence, we know that the series will converge absolutely, when $\left|x-x_0\right|<1$, that is$-1<x-x_0<1$

Implies that

role="math" localid="1649361867412" $-1+x_0<x<1+x_0$

Step 4. Thus,

Now, to find the value of $x_0$, such that the series $\sum _{k=0}^{\mathrm{\infty }} a_k{\left(x-x_0\right)}^k$convergence conditionally at both role="math" localid="1649361584817" $x=-4$and role="math" localid="1649361594209" $x=8$, the interval of convergence must satisfy

role="math" localid="1649361570464" $-1+x_0=-4\text{and}1+x_0=8$

Hence, solving for $x_0$

$x_0=2$