Prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction.

Question

I need to prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I am getting stuck midway through the inductive step.

Here is what I have:

$\forall n\geq 2$, where $n\in\mathbb{N}, P(n)$ is the statement "$\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$".

Base case:

$\sum\limits_{i=2}^{2}\frac{1}{(i-1)i}$ = $\frac{1}{2}$ = $\frac{(2-1)}{2}$ so $P(2)$ is true.

Inductive assumption: Assume $P(n)$ is true. Need to show $\sum\limits_{i=2}^{n+1}\frac{1}{(i-1)i}$ = $\frac{((n+1)-1)}{(n+1)}$.

$\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ + $\frac{1}{n^{2}}$ = $\frac{(n-1)}{n}$ + $\frac{1}{n^{2}}$

= $\frac{n(n-1) + 1}{n^{2}}$ = $\frac{n^{2}-n+1}{n^{2}}$ I got stuck here and tried working backward from $\frac{((n+1)-1)}{(n+1)}$, which gives me:

$\frac{((n+1)-1)}{(n+1)}$ = $\frac{n}{(n+1)}$

Any suggestions on what to do next or where I may be going wrong?

Answer 1

For $n\geq 2$, let $S(n)$ denote the statement $$ S(n) : \sum_{i=2}^n \frac{1}{(i-1)i}=\frac{n-1}{n}. $$ Base step ($n=2$): $S(2)$ says that $\frac{1}{(2-1)2}=\frac{1}{2}=\frac{2-1}{2}$, and this is true.

Inductive step: Fix some $k\geq 2$ and assume that $S(k)$ is true where $$ S(k) : \sum_{i=2}^k \frac{1}{(i-1)i}=\frac{k-1}{k}. $$ To be shown is that $S(k+1)$ follows where $$ S(k+1) : \sum_{i=2}^{k+1} \frac{1}{(i-1)i}=\frac{k}{k+1}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} \sum_{i=2}^{k+1} \frac{1}{(i-1)i} &= \frac{1}{k(k+1)}+\sum_{i=2}^{k}\frac{1}{(i-1)i}\tag{by defn. of $\Sigma$}\\[1em] &= \frac{1}{k(k+1)}+\frac{k-1}{k}\tag{by ind. hyp.}\\[1em] &= \frac{1+(k-1)(k+1)}{k(k+1)}\tag{common denom}\\[1em] &= \frac{k^2}{k(k+1)}\tag{simplify}\\[1em] &= \frac{k}{k+1}\tag{desired expression}, \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 2$. $\blacksquare$

Answer 2

You can try an equivalence of Principle of Induction, The Well-ordering principle. Let be $X=\{n\in\mathbb{N}:\displaystyle\sum_{i=2}^{n}\frac{1}{(i-1)i}\neq\dfrac{n-1}{n}\}$, Suposse $X\neq\emptyset$ so for the Well-ordering principle, there exist some $k\in X$ the minimum element, so $k\neq 2$ thus $2\in X^c$ (you must show this in your proccedure). $$\displaystyle\sum_{i=2}^{k}\frac{1}{(i-1)i}\neq\dfrac{k-1}{k}...(1)$$ Thus $k\neq1, \exists k-1\in\mathbb{N}$ such $s(k-1)=k$, $k-1<k$ so $k-1\in X^c$, because $k$ is the minimun of $X$: $$\displaystyle\sum_{i=2}^{n-1}\frac{1}{(i-1)i}=\dfrac{k}{k-1}...(2)$$

In $(2)$, sum to both sides $\dfrac{1}{(k-1)k}$ and with some aritmetic you must show and contradiction that holds to suppose $X\neq\emptyset$, then $X=\emptyset$ and it holds for all $n\in\mathbb{N}\setminus{1}$

Answer 3

Assume $\sum\limits_{i=2}^n\frac{1}{(i-1)i}=\frac{n-1}{n}$. Then $$\sum_{i=2}^{n+1}\frac{1}{(i-1)i}=\sum_{i=2}^n\frac{1}{(i-1)i}+\frac{1}{n(n+1)}\stackrel{\text{ind. hyp.}}=\frac{n-1}{n}+\frac{1}{n(n+1)}=$$$$=\frac{(n-1)(n+1)+1}{n(n+1)}=\frac{n^2-1+1}{n(n+1)}=\frac{n^2}{n(n+1)}=\frac{n}{n+1}=\frac{(n+1)-1}{n+1}$$

Your problem was that you added $\frac{1}{n^2}$ instead of $\frac{1}{n(n+1)}$. You want what you add to be the $n+1$'th member of the sequence $a_n=\frac{1}{(n-1)n}$ so that you find $\sum\limits_{i=2}^{n+1}a_i$ (which is $\sum\limits_{i=2}^n a_i+a_{n+1}$) given $\sum\limits_{i=2}^n a_i$.

You should also denote the sum by $\sum\limits_{i=2}^n \frac{1}{(i-1)i}$, not $\sum\limits_{i=2}^n \frac{1}{(n-1)n}$.

In addition to that, you later claimed that $\frac{n}{n+1}=n+1$, which is false.

Answer 4

You are asked to use induction, but it's not necessary.

$$\sum_{k=2}^n \frac{1}{k(k-1)} = \sum_{k=2}^n \left(\frac{1}{k-1} - \frac 1k\right) = 1 - \frac{1}{n} = \frac{n-1}{n},$$ because the sum in the middle is telescopic.