# Lim n approaches infinity# 6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=#?

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lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?

lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?

# lim_(n rarr oo) 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] = 11 #

So:

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I'm guessing this is typed incorrectly.

So here's what I guess the expression should be:

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After expanding

so

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The limit of the given expression as (n) approaches infinity is (\frac{1}{3}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*2)+1/(2*3)+...+1/(n(n+1))+...#?
- What if L'hospital's rule doesn't work?
- How do you test for convergence of #Sigma (-1)^n(1-n^2)# from #n=[1,oo)#?

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