Q. 8 - Calculus

Question

Show that $\sum _{k=0}^{\mathrm{\infty }} \frac{1}{1+2^k}{x}^k$, the power series in $x$ from Example 1, diverges when  $x=-2$

Step-by-Step Solution

Answer

Ans: The power series $\sum _{k=0}^{\mathrm{\infty }} \frac{1}{1+2^k}{x}^k$ diverges when $x=-2$.

Step 1. Given information.

given,

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

Step 2. We evaluate the series when x = - 2

  So,

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

Therefore, when $k\to \mathrm{\infty },\underset{x\to \mathrm{\infty }}{lim} \frac{(-1{)}^k{2}^k}{1+2^k}=1$

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