Q. 11 - Calculus

Question

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

Step 1. Given information.

given,

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

Step 2. Evaluate the series when x = 1 .

  So,

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

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

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

So, 

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

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

Step 4. Thus,

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