Compute the following sum

Question

I'm struggling with this problem

Compute the following sum:

$$ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times4} + \cdots +\frac{1}{99 \times 100} $$

Hint: Consider telescoping series,

Please help me.

Answer 1

Expanding on my comment above

$$\frac{1}{1\cdot 2} = \frac{1}{1} - \frac{1}{2}, \quad \frac{1}{2\cdot 3} = \frac{1}{2} - \frac{1}{3}, \quad \cdots$$

Now do you see what to do?

Added: $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{99\cdot 100} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{99} - \frac{1}{100}$$

Answer 2

There is a nice closed form for sums like this. We have $$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$$ To use the formula $\sum_{i=1}^n \frac{1}{i(i+1)}$, you first choose $n$ and then we begin summing from $i=1$ until we get to $n$. For example if $n=2$, we have $$\sum_{i=1}^2 \frac{1}{i(i+1)} = \frac{1}{1(1+1)}+\frac{1}{2(2+1)} = \frac{1}{2}+\frac{1}{6}$$ or if $n= 80$ we'd have $$\sum_{i=1}^{80} \frac{1}{i(i+1)} = \frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\ldots + \frac{1}{79(79+1)}+\frac{1}{80(80+1)}$$ It is very cumbersome to add up $80$ fractions, so this is where the first equation I mentioned is very useful. In both cases the sums I have above for $n=2$ and $n=80$ are equivalent to $\frac{n}{n+1}$. More explicitly, $$\frac{1}{2}+\frac{1}{6} = \frac{2}{2+1}$$ and $$\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\ldots + \frac{1}{79(79+1)}+\frac{1}{80(80+1)} = \frac{80}{80+1}$$ I used examples of $n=2$ and $n=80$. What value of $n$ corresponds to your problem? When you know that, simply plug it into $\frac{n}{n+1}$.