What's the limit of sequence of partial sums for the series (from 1 to ∞ ) Σ ln(n)/n^3 ?
The tools I know about for finding limits are comparison test, integral test, ratio test, nth term test, and alternating series test.
However, I can't get integral of this (can I?) and ratio doesn't seem to get anywhere. n th term test does =0 which supports convergence. I don't seem how to use comparison test. So what is this limit and how do you get it? What's the best way?
More Questions From This User See All
Answers & Comments
Verified answer
It's possible to take the integral of this. Use integration by parts.
Let our corresponding function for the series be f(x) = ln(x)/x^3. Use parts.
Let u = ln(x). dv = x^(-3) dx
du = (1/x) dx. v = (-1/2)x^(-2)
(-1/2)x^(-2)ln(x) - Integral ( (-1/2)(1/x)(x^(-2)) dx )
(-1/2)x^(-2)ln(x) + (1/2) Integral ( x^(-3) dx )
Using the reverse power rule,
(-1/2)x^(-2)ln(x) + (1/2) [ (-1/2)x^(-2) ]
(-1/2)x^(-2)ln(x) - (1/4) [ x^(-2) ]
Remember that we have to evaluate this as an improper integral; that is, evaluate it from 1 to infinty.
(-1/2)x^(-2)ln(x) - (1/4) [ x^(-2) ] {from 1 to infinity}
(-1/2)ln(x)/x^2 - (1/4) [ 1/x^2 ] {from 1 to infinity}
Which means we have to express this as a limit.
lim { (-1/2)ln(x)/x^2 - (1/4) [ 1/x^2 ] } {from 1 to t}
t -> infinity
Evaluate from 1 to t.
lim { (-1/2)ln(t)/t^2 - (1/4) [ 1/t^2 ] } - { (-1/2)ln(1)/1^2 - (1/4) [ 1/1^2 ] }
t -> infinity
lim { (-1/2)ln(t)/t^2 - (1/4) [ 1/t^2 ] } - { 0 - (1/4) }
t -> infinity
lim { (-1/2)ln(t)/t^2 - (1/4) [ 1/t^2 ] } - { -(1/4) }
t -> infinity
lim (-1/2)ln(t)/t^2 - (1/4) [ 1/t^2 ] + (1/4)
t -> infinity
As t approaches infinity, (1/t^2) approaches 0, and 1/4 is a constant so it just says the same.
lim { (-1/2)ln(t)/t^2 } - (1/4)(0) + (1/4)
t -> infinity
lim { (-1/2)ln(t)/t^2 } + (1/4)
t -> infinity
We can solve that limit using L'Hospital's rule.
lim (-1/2) [ (1/t)/(2t) ] + 1/4
t -> infinity
lim (-1/2) [ (1/(2t^2) ] + 1/4
t -> infinity
The limit approaches 0, and our final answer is
1/4
Since the integral converges, the series converges too.
*Edited for correct response*
According to the properties of the logarithmic function, for every x > 0 we have ln(x) <= x-1. So, for every n >=1 we have 0 <= ln(n) <= n-1 < n, so that 0 <= ln(n)/n^3 < n/n^3 = 1/n^2. Since the series Σ l(1/n^2) converges, it follows, by comparisson, that Σ ln(n)/n^3 converges too.
But finding its limit doesn't seem to be an easy task. I guess this only can be achieved by numerical means.
You have to use L'Hospital's rule. Take the derivative of the numerator and denominator and then simplify. Once you do that you can plug in infinity again and you will get 0. Which is the limit of this series.