The question:
- a) Differentiate both sides of the geometric series with respect to $r$: $$~~\displaystyle\sum_{i=0}^nr^i=\frac{1-r^{n+1}}{1-r}$$
- b) Use the result in part (a) to show that (Assume that $0 < r < 1$). $$\displaystyle\sum_{i=1}^nir^i < \frac{r}{(1-r)^2}$$ for all $n\ge1$.
My attempts:
- a) $$\sum_{i=0}^nir^{i-1}=\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}$$
- b) With mathematical induction: $$P(n): \sum_{i=1}^nir^i < \frac{r}{(1-r)^2}$$
$$n=1, LHS = r, RHS = \frac{r}{(1-r)^2}$$ $$LHS < RHS$$
Assume $P(k)$ is true $$P(k): \sum_{i=1}^kir^i < \frac{r}{(1-r)^2}$$
Prove $P(k+1)$ is true: $$P(k+1): \sum_{i=1}^{k+1}ir^i < \frac{r}{(1-r)^2}$$ $$\sum_{i=1}^kir^i+\sum_{i=k+1}^{k+1}ir^i < \frac{r}{(1-r)^2}$$
I am currently stuck here.