Q. 14 - Calculus

Question

 Let α ∈ R. Explain why you can find a series $\sum _{k=1}^{\mathrm{\infty }} a_k$

with all nonzero terms that converges to α. You may wish

to use your answer to Exercise 13.

Step-by-Step Solution

Answer

The geometric series$\sum _{k=1}^{\mathrm{\infty }} a_k$ with all non-zero terms that converges to $\alpha$ .

Step 1. Given information

We have been given a series that is converges to $\alpha$.

Step 2. Proving the series ∑ k = 1 ∞   a k . to be convergent .

Considering the series $\sum _{k=1}^{\mathrm{\infty }} a_k.$

The objective is to explain why the series converges to zero .

the series $\sum _{k=1}^{\mathrm{\infty }} \frac{1}{2^k}$ is a geometric series with constant ratio of $\frac{1}{2}$ which is less than 1

hence the series is convergent .

Step 3.Finding the sum of series

The series $\sum _{k=1}^{\mathrm{\infty }} a_k=\sum _{k=1}^{\mathrm{\infty }} \frac{1}{2^k}$ converges to the sum

$\begin{array}{r}=\frac{\frac{1}{2}}{\frac{1}{2}}\\ =1\end{array}$

The series converges $\sum _{k=1}^{\mathrm{\infty }} a_k=\sum _{k=1}^{\mathrm{\infty }} \frac{1}{2^k}$ the sum of 1 .

Step 4. Finding the explanation of problem

The series $\sum _{k=1}^{\mathrm{\infty }} a_k=\sum _{k=1}^{\mathrm{\infty }} \frac{1}{2^k}$ converges to 1.

If each term of the series is multiplied by constant the series become $\sum _{k=1}^{\mathrm{\infty }} \alpha a_k=\sum _{k=1}^{\mathrm{\infty }} \frac{\alpha }{2^k}$

The convergence of the series changes from $1$ to $\alpha$

The geometric series $\sum _{k=1}^{\mathrm{\infty }} a_k$ with all non-zero terms that converges to $\alpha$.