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How can I get [if it's possible]: $$ 2- \frac{1}{n+1} $$ from this: $$2- \frac{1}{n} + \frac{1}{n(n+1)} $$

EDIT:

I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$

How do I proceed from here?

Hodaya Shalom
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    I assume you wish to ask how to prove this equality?

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

     – PrincessEev Sep 03 '19 at 06:53
  • @EeveeTrainer - yes – Hodaya Shalom Sep 03 '19 at 06:54
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    I would start from the term $\frac{1}{n(n+1)}$. Try to write it in a form $$ \frac{1}{n(n+1)} \equiv \frac{A}{n} + \frac{B}{n+1} $$ where $\equiv$ means that they are algebraically equivalent. Can you find $A$ and $B$ so that this is true? Of course we need to assume $n\neq 1$ and $n\neq 0$. Hint: multiply both sides of the equation by $n(n+1)$. – Matti P. Sep 03 '19 at 06:56
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    I'd forget the $2$s and put $$-\frac1n+\frac1{n(n+1)}$$ over a common denominator. – Angina Seng Sep 03 '19 at 06:57
  • A common trick is $$\frac{1}{n(n+1)} = \frac{(n+1)-n}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ Similarly, one has $$\frac{1}{n(n+1)(n+2)} = \frac12 \frac{(n+2) - n}{n(n+1)(n+2)} = \frac12\left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right]$$ This sort is trick is pretty useful when you turn a sum into a telescoping one... – achille hui Sep 03 '19 at 07:19

2 Answers2

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I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$

The mistake happened here:

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

because you forgot the minus sign. You can rewrite the left hand side as

$$2 +\frac{-n-1}{n(n+1)} + \frac{1}{n(n+1)}$$ and continue from there.

5xum
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There is no need to care about the term $2$, it suffices to establish

$$\frac1{n+1}=\frac1n-\frac1{n(n+1)}.$$

If you multiply by $n(n+1)$, this is

$$n=n+1-1.$$