Prove by induction help? I'm trying to study for a test and one of the practise questions is very confusing and not sure what to do:
Prove by induction that $$\sum_{i=1}^n \frac{3}{4^i} < 1$$ for all $n \geq 2$
The furthest I'm able to get is getting rid of the summation.
So I get $(3/4)^i n< 1$ for all $n \geq2$.
Any help would be greatly appreciated. Thanks.
 A: It is defeating you because
$${3\over 4} + {9\over 16} = {21\over 16} > 1$$
A: Prove that it's true for $n=2$.
Then prove that, if it is true for $n=k$, then it will be true for $n=k+1$. 
That's the inductive process.
A: Assuming the question is supposed to be $\displaystyle\sum_{i=1}^n \left(\dfrac{3}{4}\right)^i > 1, \ \ \forall n \geq 2$ it's pretty straight-forward to do.
Clearly it's true for $n=2$ as we get $\displaystyle\sum_{i=1}^2 \left(\frac{3}{4}\right)^i =  \frac{3}{4} + \frac{9}{16} = \frac{21}{16} > 1$
Assume then, for $n=k$ we have that $\displaystyle\sum_{i=1}^k \left(\frac{3}{4}\right)^i > 1$
Then for $n=k+1$
$\displaystyle\sum_{i=1}^{k+1} \left(\frac{3}{4}\right)^i = \displaystyle\sum_{i=1}^k \left(\frac{3}{4}\right)^i + \left(\frac{3}{4}\right)^k > 1 + \left(\frac{3}{4}\right)^k > 1 $
EDIT: I just thought I'd add something, I don't know if you just typed the question out incorrectly or if you really have been trying to show that the sum is $<1$, but it's very important to show it is at least true for one value of $n$ in the given range (so in this case any $n \geq 2$) because if you don't do this, the proof (by induction) doesn't really work. The power of showing it's true for $n=2$, means that once you've proven it by induction, it's also true for $n=3,4,5,6...$ because of the way in which you assume it's true for some $k$ and prove it's true for $k+1$.
