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Show that $$\frac1{1\cdot2}+\frac1{2\cdot3}+\frac1{3\cdot4}+\cdots+\frac1{(n-1)\cdot n}=\frac{n-1}{n}.$$

I’m having a really hard time with this question - I can’t start it with one because you can’t divide by zero, and as I go further along I still am struggling. I started with two, so I did the base step, and the induction hypothesis step, how do you prove that K+1 is true? How do I do this process and what’s the answer?

M. Overby
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    You should start by $n=2$. – hamam_Abdallah Apr 29 '20 at 02:02
  • Well, whats the first instance? It's $$\frac 1{1\cdot 2}=\frac 1{2}$$. ANd what's the second? It's $$\frac 1{1\cdot 2}+\frac 1{2\cdot 3}= \frac 23$$. So why start at $n=1$ which gives you $\frac 1{0}$ which isn't a case. Start and the first thing that IS a case, not anything that isn't. Start and $\frac 1{1\cdot 2} = \frac 12$.... which is ... $n = 2$. – fleablood Apr 29 '20 at 02:18
  • And If $\frac 1{12} + ....... + \frac 1{(k-1)k} = \frac {k-1}k$ then $\frac 1{12} + ....... + \frac 1{(k-1)k} + \frac 1{k(k+1)} = \frac {k-1}k + \frac 1{k(k+1)}$ so..... – fleablood Apr 29 '20 at 02:20
  • There is no problem with $n=1$. Then the LHS is an empty sum and the RHS is zero. – Angina Seng Apr 29 '20 at 02:28

1 Answers1

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Other approach

$$\frac{1}{1.2}=\color{green}{\frac{1}{1}}-\color{red}{\frac{1}{2}}$$

$$\frac{1}{2.3}=\color{red}{\frac{1}{2}}-\frac{1}{3}$$

$$\frac{1}{3.4}=\frac 13 -\frac 14$$ ...

$$\frac{1}{(n-1)n}=\frac{1}{n-1}-\color{green}{\frac{1}{n}}$$

if we sum, we find after telescoping $$1-\frac 1n=\frac{n-1}{n}$$

hamam_Abdallah
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