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- anonymous

limits
find the limit as n->infinity of (1/n+1) + (1/n+2) + ... + (1/2n)

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- anonymous

limits
find the limit as n->infinity of (1/n+1) + (1/n+2) + ... + (1/2n)

- jamiebookeater

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- anonymous

Wouldn't the fractions as n became infinitely large basically be equal to 1 over some really big number (which basically means the fraction gets closer and closer to zero)? So the limit would be zero since you would be adding a bunch of zeros together right?

- anonymous

yeah thats how i reasoned also...
i just had an exam and that was a problem so i was seeing if other people knew any other ways to solve so i could kinda check myself

- anonymous

well an easy way to look at limits especially as N approaches infinity is to look at the numerator and denominator in relation to where the infinity is. If it is on the top then the limit is infinity assuming the bottom isn't over infinity in which case the limit would be one. If the infinity is on the bottom then its really just some number over a really big number (ie 1/199999999999999999999999999) which is something like 0.0000000000 yada yada

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- anonymous

yeah...idk my profesor was saying that we needed to know explicit representation for it...like as a hint.
but the series of n numbers is n(n+1)/2 explicitly...and so i was like trying to reciprocate but that would be incorrect since you start at 1/(n+1)
and so i ended up like scribbling a few inequalities and then concluding that the limit was zero haha

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