Q. 12 - Calculus

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?

 

Step-by-Step Solution

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 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 $[-1,1 ]$.