Question

Show that the power series $\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}{x}^k$converges absolutely when $x=1$and when $x=-1$. What does this behavior tell you about the interval of convergence for the series?

Short answer

Ans: The power series $\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}{x}^k$has the interval of convergence $[-1,1]$

Step-by-step solution

Step 1. Given information.

given,

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

Step 2. Evaluate the series when x=1

So,

$\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}{x}^k=\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}(1{)}^k\\ =\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}\end{array}$

So, for $x=1$, we have the alternating harmonic series which converges conditionally.

Step 3. We evaluate the series when x=-1

So,

$\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}{x}^k=\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}(-1{)}^k\\ =\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^{2k}}{k^2}\\ =\sum _{k=1}^{\mathrm{\infty }} \frac{1}{k^2}\end{array}$

So, for $x=-1$, we have the alternating harmonic series which converges conditionally.

Step 4. Thus, 

Therefore, the power series$\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k}{k^2}{x}^k$ has the interval of convergence$[-1,1]$.