How can I prove that n/(n+1) = 1-[1/(n+1)]?

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To prove that n/(n+1) equals 1 - [1/(n+1)], one approach is to rewrite n/(n+1) as ((n+1) - 1)/(n+1). This manipulation reveals that the expression simplifies to 1 - [1/(n+1)]. Alternatively, starting with the second fraction, 1 - [1/(n+1)], and finding a common denominator can also lead to the same result. This method emphasizes the importance of understanding how to manipulate fractions to demonstrate equivalence. Overall, both strategies highlight the relationship between the two expressions effectively.
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I am beginning to teach myself series/sequences but I keep running into issues getting started.

A common example I keep seeing is: n/(n+1)
This almost always gets set to: 1 - [1/(n+1)]

I have tried to work out the proof of this myself, but I can't seem to get the two to equal each other/manipulate the first to equal the second.

Could someone point me towards what i have to do to the first fraction? Should be an easy thing but I can't put my finger on it.
 
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Hey,

write n/(n+1) as ((n+1)-1)/(n+1)

What do you see?
 
ooohhhhhhh
thanks a lot, very helpful, good hint.
 
red_hockey said:
I have tried to work out the proof of this myself, but I can't seem to get the two to equal each other/manipulate the first to equal the second.

Could someone point me towards what i have to do to the first fraction? Should be an easy thing but I can't put my finger on it.

The easier way of proving this would be to work with the second fraction, then hopefully come up with how you can reverse the procedure.
Why we want to work with the end result is because the first fraction is just one fraction, which means to get from

1-\frac{1}{n+1} to \frac{n}{n+1} we need to start by taking the common denominator, that is, putting the fraction all under the same denominator.
 

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