Tuesday, 19 July 2016

calculus - Proving the convergence of suminftyn=2frac1n(lnn)2



I am attempting to prove if the below expression converges or diverges:




n=21n(lnn)2



I decided to try using the limit convergence test. So then, I set f(n)=1n(lnn)2 and g(n)=1n and did the following:



lim



I know that \sum \frac{1}{n} diverges due to properties of harmonic series, and so concluded that my first expression \frac{1}{n (\ln n)^2} also must diverge.



However, according to Wolfram Alpha, and as suggested by this math.SE question, the expression actually converges.



What did I do wrong?


Answer



In the limit comparison test, \sum_n f(n) and \sum_n g(n) both will behave the same iff
\lim_{n \to \infty} \dfrac{f(n)}{g(n)} = c \in (0,\infty)

In your case, the limit is 0 and hence you cannot conclude anything.


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