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Use Mathematical induction to prove that for all integers, $n$ is greater than or equal to $1$. I am confused on what to do after I do the the basis step that is using $n$ as $1$.

$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$$

Martin Sleziak
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alan
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    From your title and text, it sounds like you want to prove $\forall n\in\Bbb N\space n\geq 1$. It seems that what you actually want to prove is that the formula at the end of your question is true for all $n$? – Jack M Apr 06 '14 at 04:02
  • The next term to be added is $1/(n+1)((n+1)+1)$, which does equal the difference between the right-hand sides, "after" $1-1/((n+1)+1)$ and "before" $1-1/(n+1)$. –  Apr 09 '15 at 08:32

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Let $\displaystyle f(n):\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{n(n+1)}=1-\frac1{n+1}$

holds true for $n=m$

$\displaystyle\implies f(m):\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{m(m+1)}=1-\frac1{m+1}\ \ \ \ (1)$

For $n=m+1$ adding $\displaystyle\frac1{(m+1)(m+2)}$ to both sides of $(1),$ $\displaystyle\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{m(m+1)}+\frac1{(m+1)(m+2)}=1-\frac1{m+1}+\frac1{(m+1)(m+2)}$

Now $\displaystyle\frac1{(m+1)(m+2)}=\frac{(m+2)-(m+1)}{(m+1)(m+2)}=\frac1{m+1}-\frac1{m+2}$

or $\displaystyle-\frac1{m+1}+\frac1{(m+1)(m+2)}=\frac{-(m+2)+1}{(m+1)(m+2)}=\cdots$

Can you take it home from here?

Don't forget to demonstrate the base case

lab bhattacharjee
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