Question

Complete Example 4 by showing that the power series $\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}(2x-3{)}^k$diverges when $x=2$.

Short answer

Ans: By the divergence test, the power series $\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}(2x-3{)}^k$diverges when $x=2$

Step-by-step solution

Step 1. Given information.

given,

$\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}(2x-3{)}^k$

Step 2. We evaluate the series when x=2

So,

$\begin{array}{r}\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}(2x-3{)}^k=\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}[2\left(2\right)-3{]}^k\\ =\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}\end{array}$

Therefore,$k\to \infty$,$\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}\to \infty$

Step 3. Thus,

Thus, by the divergence test, the power series $\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k\frac{k}{3k+1}(2x-3{)}^k$diverges when $x=2$