Q. 20 - Calculus

Question

Let $\alpha \text{ and }\beta$ be two distinct positive numbers less than $1$. Explain why the ratio test cannot be used on the series $\alpha +\beta +{\alpha }^2+{\beta }^2+{\alpha }^3+{\beta }^3+\cdots$ . Then show that the series converges and find its sum.

Step-by-Step Solution

Answer

Hence proved.

Step 1, Given information.

The given series is $\alpha +\beta +{\alpha }^2+{\beta }^2+{\alpha }^3+{\beta }^3+\cdots$

Step 2. Conclusion.

We can rewrite the series as,

$\left(\alpha +{\alpha }^2+{\alpha }^3\dots \right)+\left(\beta +{\beta }^2+{\beta }^3+\dots \right)$

$$\begin{gathered} \text{ Rewrite }\alpha \left(1+\alpha +{\alpha }^2\dots \right)=\alpha \sum _{k=0}^{\infty }\frac{1}{{\alpha }^k} \\ \text{ And }\beta \left(1+\beta +{\beta }^2+\dots \right)=\beta \sum _{k=0}^{\infty }\frac{1}{{\beta }^k} \\ Therefore, \\ \left(\alpha +{\alpha }^2+{\alpha }^3\dots \right)+\left(\beta +{\beta }^2+{\beta }^3+\dots \right)=\alpha \sum _{k=0}^{\infty }\frac{1}{{\alpha }^k}+\beta \sum _{k=0}^{\infty }\frac{1}{{\beta }^k} \\ Hence, the series converges. \\ Now, the sum is, \\ \alpha \sum _{k=0}^{\infty }\frac{1}{{\alpha }^k}+\beta \sum _{k=0}^{\infty }\frac{1}{{\beta }^k}\text{ is }\alpha \left(\frac{1}{1-\alpha }\right)+\beta \left(\frac{1}{1-\beta }\right) \\ \alpha \left(\frac{1}{1-\alpha }\right)+\beta \left(\frac{1}{1-\beta }\right)=\left(\frac{\alpha }{1-\alpha }\right)+\left(\frac{\beta }{1-\beta }\right) \end{gathered}$$