# Calculus: Determine the convergence/divergence of the series

n=1 to infinity Σ [ ln(n+1)*(-1)n+1/(n+1) ]

ok, this is a simple application of the alternating series convergence criteria: http://en.wikipedia...ernating_series_test
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the point is that we need to prove before that:
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the sequence a(n)=ln(n)/n (for n>=2) is monotonically decreasing
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and that the limit for n->infinity of a(n) is 0
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the limit is a simple application of the fact that the logarithm goes to infinity of a smaller order than the infinity of n, therefore the limit of a(n) is 0
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the monotonically decreasing part is a little more elaborate
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calculate d(n)=a(n+1)-a(n)
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through some algebra you get that:
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d(n)=log((n+1)^n/n^(n+1))/(n*(n+1))
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now, the denominator is always positive, so let's check the sign of the numerator as n varies (n>=2)
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let's check therefore the argument of the logarithm
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if it is greater than 1, then the numerator will be positive, otherwise if it is between 0 and 1 it will be negative
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((n+1)^n)/(n^(n+1))=(1+1/n)^n*(1/n)
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now we know that (1+1/n)^n
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and e=2.712..... therefore e
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for n=2 , we get that (1+1/2)^2 / 2 = 9/8>1 and therefore the numerator is positive
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but from n>=3 we get that (1+1/n)^n / n
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therefore the series is monotonically decreasing for n>=3 (which means, since the original series was in terms of n+1, from n>=2)
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the first term is a(1) and it is finite, the partial sum from n>=2 is an alternating series of a monotonically decreasing series of positive terms, therefore the alternating series criterium applies
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Alternative proof, in case you don't want to use the alternating series criterium
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the series is equivalent to sum from 2 to infinity of ln(n+1)/(n+1)-ln(n)/n