2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$
  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.

  4. As it turns out, the stricter inequality (or even equality) is much easier to prove. Proe by induction is straightforward.

  5. Since $1-1/4^n<1$ you also obtain the originally desired result.

$\endgroup$
  • $\begingroup$ Thank you and you are correct about your first comment. $\endgroup$ – Michel Tamer Nov 21 '13 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.