what test could we use in this series to find if it converges or not...what test is the simpliest?

Question

oo
E 3ln(2k) / k^2
k=1

i was thinking of the comparison test...would that work and how?

Answer 1

Have you learnt Ermakoff's Test? I think it would work well with this question.

∑f(x) where f(x) is monotonic non-increasing will converge if
lim(x→∞) e^x f(e^x)/f(x) < 1 and diverge if greater than 1.


So ∑3ln{2k}/k^2 will converge if
lim(k→∞) e^k [3ln(2e^k) / (e^k)^2] / [3ln(2k) / k^2]
= lim(k→∞) [e^k (ln[2] + k) k^2] / [(ln(2) +ln(k)) e^2k]
= lim(k→∞) [ 3(ln[2] + k) k^2] / [3(ln(2) +ln(k)) e^k]
< lim(k→∞) [ (k + k)k^2 / [(ln(2) + ln(2)) e^k]
As ln(2) < k
= lim(k→∞) [ 2k^3 / (2ln2 e^k) ]
= 1/ln2 lim(k→∞) [ k^3 / e^k ]
Now this converges as k^3 is polynomic while e^k is exponential.
So by Ermakoff's Test it converges.

Answer 2

Another solution: For k>2, the terms ln(k)//k^2 are monotonically decreasing. So, we know Sum ln(k)//k^2 converges if and only if Sum (2^n ln(2^n))/((2^n)^2) = Sum (2^n n ln(2))/((2^n)^2) =Sum ln(2) (n/2^n) converges.

Since lim [(n+1)/2^(n+1)]/[n/2^n] = lim (n+1)/n * lim (2^n/2^(n+1)) = 1 * 1/2 = 1/2 < 1, it follows Sum ln(2) (n/2^n) converges and, therefore, Sum(ln(k)/k^2 converges. Since ln(2k)/k^2 = ln(2)/ k^2 + ln(k)/k^2 and Sum 1/k^2 converges, it follows Sum ln(2k)/k^2 converges and your series converges to 3 times its limit.

Answer 3

It only depends on whether the numerator grows faster than the denominator.
Numerator : d (3ln(2k)) / dk = 6/k
Denominator : d( k^2 ) / dk = 2k

Denominator obviously grows faster than numerator, so the function will converge.