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.

Short answer

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

Step-by-step solution

Step 1. Given information 

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

Step 2. Proving the series ∑k=1∞ ak. 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$\alpha$.