Q. 66 - Calculus

Question

Use the principle of mathematical induction to prove that if $a_{k+1}<a_{k}r$ for every k ≥ N, then $a_{N+n}<a_N{r}^n$. Proving this implication completes our proof of the ratio test.

Step-by-Step Solution

Answer

Hence, proved.

Step 1. Given Information.

Given $a_{k+1}<a_{k}r for every k⩾N.$

Step 2. Proof.

Let suppose it is true for k=N, then 

$a_{N+1}<a_{N}r.$

For k=N+1, we get:

$$\begin{gathered} a_{N+2}<a_{N+1}r, \\ {a}_{N+2}<\left(a_{N}r\right)r \\ {a}_{N+2}<a_N{r}^2. \end{gathered}$$

The result is true for k=N.

Assume that the result is true for k=N+n-1,

$a_{N+n}<a_N{r}^n.$

Now we have to prove it for k=N+n.

Step 3. Proof part 2.

$$\begin{gathered} For k=N+n, \\ {a}_{N+n+1}<a_{N+n}r, \\ {a}_{N+n+1}<\left(a_N{r}^n\right)r, from step 2. \\ {a}_{N+n+1}<a_N{r}^{n+1}, \\ Hence, the result is true for k=N+n. \\ Hence by the principle of mathematical induction, result is true for \\ every positive numbers k⩾N. \end{gathered}$$