Proof by induction - fractions

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SUMMARY

The forum discussion centers on proving the equation $$\sum_{i=1}^{n}\frac{1}{i(i+1)}=1-\frac{1}{(n+1)}$$ using mathematical induction. The user initially struggled to manipulate the expression to reach the desired form of $$1-\frac{1}{k+2}$$. The breakthrough came when the user realized that separating the fraction $$\frac{1}{i(i+1)}$$ into $$\frac{1}{i} - \frac{1}{i+1}$$ simplifies the proof process significantly.

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mikky05v
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Homework Statement


I have been working on this proof for a few hours and I can not make it work out.

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

i need to get to
$$1-\frac{1}{k+2}$$

I get as far as
$$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
then I have tried
$$1-\frac{(k+2)+1}{(k+1)(k+2)}$$
by multiplying the left fraction by (k+2) which got me nowhere.

What am I doing wrong?
 
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disregard, I figured it out. I simply had to separate 1/i(i+1) into 1/i - 1/(i+1)
 

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