Calculus: Determine the convergence/divergence of the series

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

Answers
ok, this is a simple application of the alternating series convergence criteria: http://en.wikipedia...ernating_series_test
advmc
30 January 2012
the point is that we need to prove before that:
advmc
30 January 2012
the sequence a(n)=ln(n)/n (for n>=2) is monotonically decreasing
advmc
30 January 2012
and that the limit for n->infinity of a(n) is 0
advmc
30 January 2012
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
advmc
30 January 2012
the monotonically decreasing part is a little more elaborate
advmc
30 January 2012
calculate d(n)=a(n+1)-a(n)
advmc
30 January 2012
through some algebra you get that:
advmc
30 January 2012
d(n)=log((n+1)^n/n^(n+1))/(n*(n+1))
advmc
30 January 2012
now, the denominator is always positive, so let's check the sign of the numerator as n varies (n>=2)
advmc
30 January 2012
let's check therefore the argument of the logarithm
advmc
30 January 2012
if it is greater than 1, then the numerator will be positive, otherwise if it is between 0 and 1 it will be negative
advmc
30 January 2012
((n+1)^n)/(n^(n+1))=(1+1/n)^n*(1/n)
advmc
30 January 2012
now we know that (1+1/n)^n
advmc
30 January 2012
and e=2.712..... therefore e
advmc
30 January 2012
for n=2 , we get that (1+1/2)^2 / 2 = 9/8>1 and therefore the numerator is positive
advmc
30 January 2012
but from n>=3 we get that (1+1/n)^n / n
advmc
30 January 2012
therefore the series is monotonically decreasing for n>=3 (which means, since the original series was in terms of n+1, from n>=2)
advmc
30 January 2012
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
advmc
30 January 2012
Alternative proof, in case you don't want to use the alternating series criterium
advmc
30 January 2012
the series is equivalent to sum from 2 to infinity of ln(n+1)/(n+1)-ln(n)/n
advmc
30 January 2012
doing some algebra, the single term in the series is
advmc
30 January 2012
equal to: log ( (1+1/n)^n1/n)) / (n(n+1))
advmc
30 January 2012
nevermind, it's getting too long, try to understand the wiki link to the proof of the alternating series criterium and use the first one
advmc
30 January 2012
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