# Prove a summation inequality by induction: $\sum_{i=1}^n \frac{3}{4^i} < 1$

I was having trouble proving by induction with this problem. $$\sum_{i=1}^n \frac{3}{4^i} < 1$$ for all $n \geq 2$ I went to see my professor and he said try proving this equality $$\sum_{i=1}^n \frac{3}{4^i} < 1 - 1/4^n$$ Where did he get the $$1-(1/4^n)$$ from? How would I prove this? And is it still proving the same inequality?

1. The "improved" inequality is wrong as stated, it should be $\le$ (or even $=$) instead of $<$
2. You can hardly use induction with the original inequality. If you only have $s_n<1$, you cannot conclude that $s_{n+1}<1$ because you always have $s_{n+1}>s_n$. In other words, you need that $s_n$ is sufficiently smaller than $1$ (and need to show that $s_{n+1}$ is not just smaller, but sufficiently smaller than $1$)
3. You might get the $1-1/4^n$ from looking at the first few sums ($\frac34$, $\frac{15}{16}$, $\frac{63}{64}$) and smelling the pattern.
5. Since $1-1/4^n<1$ you also obtain the originally desired result.