Q. 10 - Calculus

Question

Show that the power series $\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{2k+1}{x}^{2k+1}$ converges conditionally 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 $\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{2k+1}{x}^{2k+1}$ has the interval of convergence $[-1,1 ]$

Step 1. Given information.

given,

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

Step 2. Evaluate the series when x = 1

  so,

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

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

Step 3. Evaluate the series when x = - 1

 So,

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

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

Step 4. Thus,

 Therefore, the power series  $\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{2k+1}{x}^{2k+1}$ has the interval of convergence $[-1,1 ]$.