prove using induction proving method

Question

$$ \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1} $$

I was wondering how to solve this equation using induction method, I am not quite sure where to begin as I am new to proving techniques.

Check if it's correct when n=1. Then, assume it's correct when for n=k, prove it's correct for n=k+1 as well. — Divide1918, Dec 16 '20 at 03:31
Does this answer your question? How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$ — GNUSupporter 8964民主女神地下教會, Dec 16 '20 at 10:26
Please use https://approach0.xyz to search whether your question has been asked first. — GNUSupporter 8964民主女神地下教會, Dec 16 '20 at 10:27

score 2 · Accepted Answer · answered Dec 16 '20 at 03:37

Firstly, write your equation in summation notation. It makes it easier to work with in my experience.

$$\sum_{k=1}^n\frac{1}{k(k+1)}=1-\frac{1}{n+1}\tag 1$$

Now, the induction proof is done in three steps. Firstly, enumerate the base case $n=1$ $$\frac{1}{1\cdot 2}=\frac12=1-\frac12$$ so it is true.

Second, assume it true for a given $r\in \Bbb N$: $$\sum_{k=1}^r\frac{1}{k(k+1)}=1-\frac{1}{r+1}\tag 2$$ and use this to prove it true for $n=r+1$, i.e. that $(2)$ implies: $$\sum_{k=1}^{r+1}\frac{1}{k(k+1)}=1-\frac{1}{r+2}$$

To do this, add the $(r+1)$th term to the sum of $(2)$.

Then conclude your proof by induction, explaining that having showed the statement true for the base case $n=1$, and the $n+1$th case when the $n$th case has been assumed, so it is true for all $n\in \Bbb N$

prove using induction proving method

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