Closed form of a sum

Question

I am trying to derive the closed form of the sum $\sum\limits_{i=2}^n \frac{1}{i(i-1)}$ which Mathematica tells me is $\frac{n-1}{n}$. I am completely baffled on how to arrive at this result. The furthest I can get is $\sum\limits_{i=2}^n \frac{1}{i - 1} - \sum\limits_{i=2}^n \frac{1}{i}$. Is there a different way I should approach this? Any guidance would be greatly appreciated.

Context: I'm trying to prove via induction that $\sum\limits_{i=2}^n \frac{1}{i(i-1)} < 1$ for all $n \geq 2$. I can't figure out an alternate way to prove this without a closed form for the sum.

Answer 1

Hint: You are on the right track and are almost there. If you reindex the second sum, giving $$\sum_{i=2}^n \frac{1}{i} = \sum_{j=3}^{n+1} \frac{1}{j-1},$$ do you see how the sums collapse?

Answer 2

Hint: Most of the terms in $\sum_{i=2}^n 1/(i-1)$ and $\sum_{i=2}^n 1/i$ are the same. Write out a few examples for small $n$ and you'll see how it works.

Answer 3

$\frac{1}{2\times 1}+\frac{1}{3\times 2}+...+\frac{1}{n\times (n-1)}$

$=(1-\frac12)+(\frac12-\frac13)+...+(\frac{1}{n-1}-\frac{1}{n})$

$=1+(-\frac12+\frac12)+(-\frac13+\frac13)+...+(-\frac{1}{n-1}+\frac{1}{n-1})-\frac1n$

$=1-\frac{1}{n}$

$=\frac{n-1}{n}$