Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$ 
If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then
  $$S\gt 1,$$

but how?
I understood that there are $451$ pair of terms. So clubbed two terms together.
$\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$
But I am not able to solve it further. Such a tricky question for me.

numbers from $\frac{1}{100}$ to $\frac{1}{200}$ are larger  than $\frac{1}{200}$
and there are $100$ of them (101 actually, but doesn't matter) so their sum is larger than $100\times \frac{1}{200}=\frac12$
You can repeat a similar argument to show that from $\frac{1}{200}$ to $\frac{1}{300}$ the sum is larger than $\frac13$
until you arrive to numbers from $\frac{1}{900}$ to $\frac{1}{1000}$ which have sum larger than $\frac{1}{10}$
so the numbers  from $\frac{1}{100}$ to $\frac{1}{1000}$ are less than $\frac12+\frac13+\ldots+\frac{1}{10}\approx 1.93 $
Our sum is less than the actual sum ($\approx 2.308$) and is larger than $1$ so we have proved what was required to prove
hope this helps
