Find the sum of the following series: $ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$

Question

Find the sum of the following series $$ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$$

My Attempt:

$$\frac{1}{2 \cdot 3} =\frac{1}{2}-\frac{1}{3}.$$

So we can write question as:

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\\ =\frac{1}{1}-\frac{1}{101}\\ =\frac{100}{101}.$$

Am i right?

Minor note: your reasoning will be more clear and convincing if you write $$\frac1{n(n+1)}=\frac1n-\frac1{n+1}$$ instead of just the case for $1/(2\cdot3)$. — Théophile, Apr 19 '17 at 13:10
See What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{n(n+1)}$ and other questions linked there. — Martin Sleziak, Apr 19 '17 at 13:31

score 0 · Answer 1 · answered Apr 19 '17 at 14:08

_Sum up of everybody else's comments: _Yes, it is correct. The only thing I'd say is that as Theophile said, you need to make your prove more convincing, and back up every single hole in. A proof is never at an optimal proving state, but you can maximise this, by filling in every single statement you make with a law or rule, or in other cases, another statement. In the end, all the loose bits would be tied up, and your proof will be complete.

Find the sum of the following series: $ \frac{1}{1 \cdot 2}\,+\,\frac{1}{2 \cdot 3}\,+\,\frac{1}{3 \cdot 4} +\cdots\,+\frac{1}{100 \cdot 101}$

My Attempt:

1 Answers1