How to proof that $\forall n \ge 1: \sum_{i=1}^n (i(i+1))^{-1} = n(n+1)^{-1}$ using mathematical induction

Question

So for a better presentation: https://i.stack.imgur.com/HCzft.jpg

I need to prove this using mathematical induction, so i started with a base of $n=1$

Since $i$ starts at $i = 1$ I got this:

$$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{2} = \frac{1}{2}$$

So since this was true I went to the induction part: I chose $n = k$ and assumed this was true:

$$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{i(i+1)} = \frac{k}{k+1}$$

but I got stuck here since I didn't know if I needed to increment $i$ or not and because I didn't know if I needed to add the summation symbol or not and needed to do a few things with it.

Answer 1

If $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}.$$ then $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}=$$ $$=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2}$$ and since the base of an induction you made already, we are done!

Answer 2

If you want to use induction, what you should suppose is:

$A_n=\sum_{i=1}^n\frac{1}{i(i+1)}=\frac{n}{n+1}$

And using it, what you need to prove is:

$A_{n+1}=\sum_{i=1}^{n+1}\frac{1}{i(i+1)}=\frac{n+1}{n+2}$

If you observe that $A_{n+1}=A_n+\frac{1}{(n+1)(n+2)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}$, you just have to reduce to the same denominator and factorize the numerator...

Answer 3

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

Suppose that the above is true for some $n$, then for $n+1$ we have:

$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{1}{(n+1)(n+2)} + \sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{1}{(n+1)(n+2)} +\frac{n}{n+1}$$

Simplifying we get:

$$\frac{1}{(n+1)(n+2)} +\frac{n}{n+1} = \frac{n^2 + 2n + 1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)} = \frac{n+1}{n+2}$$

So if $\displaystyle \sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$ is true, then so is $\displaystyle \sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$. Simply plug in $n = 1$ to prove the base case and you're done.

Answer 4

Let's have $S_n=\sum_{i=1}^{n}\frac{1}{i(i+1)}$

$S_{n+1}=S_n+\frac{1}{(n+1)(n+2)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}$

$S_{n+1}=\frac{n(n+2)+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}=\frac{n+1}{n+2}$

Answer 5

Your assumption must be that: $$\sum_{i=1}^k\frac{1}{i(i+1)}=\frac{k}{k+1}$$

Based on that you must prove that:$$\sum_{i=1}^{k+1}\frac{1}{i(i+1)}=\frac{k+1}{(k+1)+1}$$or equivalently that $$\sum_{i=1}^{k}\frac{1}{i(i+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$$

Answer 6

$$\begin{align} \sum_{i=1}^n \frac 1{i(i+1)} &=\sum_{i=1}^n \frac {i^2-(i^2-1)}{i(i+1)}\\ &=\sum_{i=1}^n \frac i{i+1}-\frac {(i+1)(i-1)}{i(i+1)}\\ &=\sum_{i=1}^n \frac i{i+1}-\frac {i-1}i\\ &=\color{red}{\frac n{n+1}} &&\text{(by telescoping)}\end{align}$$