Sequences and Series, stuck :|?

Question

Hi everyone, I'm currently enrolled in calculus 1b and we're doing sequences and series. The following question isn't that bad, but I'm stuck. The problem is this:

(Sigma) from n=1 to infinity of ln ( n / (2n+5) ) .

I took the limit of ln ( n / (2n+5) ) and found out that it does equal 0 so by the divergency test, this series does not diverge (and therefore converges). I'm just not sure what I do from here, partial sums? telescoping series stuff? Let me know, thanks in advance!

Answer 1

Huh? The series does not converge to 0, so the sum is divergent:

Lim n->inf ( n / (2n+5) ) = Lim n->inf ( 1 / (2+5/n) ) is clearly 1/2
Lim n->inf ln ( n / (2n+5) ) = ln (1/2) = -ln(2) ~= -0.693

=> (Sigma)_n=1_inf ln ( n / (2n+5) ) does not exist

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In response to your mail asking why
Lim n->inf ( n / (2n+5) ) is not indeterminate and therefore would require De L'Hospital's Rule:

You can extract a common factor n:
Lim n->inf ( n / (2n+5) ) = Lim n->inf ( 1 / (2+5/n) ) = 1/2

Doing what you did neglects that:
Lim n->inf ln ( n / (2n+5) ) != Lim n->inf [ ln (n) - ln (2n+5) ]
= Inf - Inf = undefined

Answer 2

You did something screwy with your limit test. In the limit, (n/(2n+5) ) goes to (1/2) so in the limit =:

(Sigma) from n=1 to k of ln ( n / (2n+5) ) approaches k ln(1/2) which gives a divergent series.

Answer 3

Since the sum of logs is the log of products...see if exponentiating both sides helps:

Let S = the sum you're looking for,
then
exp(S) = exp(sigma..) = product n/(2n+5) for n=1 to inf

The numerator of that product is 1*2*3*....*inf

the denominator is 7*9*11*13*...inf

looks like sum judicious cancelling might help

Just some ideas...