I have to prove that the series $\displaystyle\sum_{n=1}^\infty {\ln(n^2) \over n^2 }$ converges. The ratio test is inconclusive, so I should use the comparison test, but which series should I compare it with? I tried ${1\over n}$, ${1 \over n^2 }$, but I need a bigger series which converges to prove that this one converges.
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2First, $\ln(n^2)= 2\ln(n)$. – Ben Grossmann Apr 24 '16 at 13:19
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2You can note, for example, that $$\ln(n)/n \leq 1/\sqrt n$$ for sufficiently large $n$ – Ben Grossmann Apr 24 '16 at 13:21
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Prove that $0\le \ln n \le \sqrt{n}$ first, and $$ 0\le \frac{\ln n^2}{n^2} =\frac{2\ln n}{n^2} \le \frac{2}{n\sqrt{n}}. $$ $\sum_{n=1}^{\infty}\frac{2}{n\sqrt{n}}$ converges.
choco_addicted
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Consider $a_n:=\frac{\ln (n^2)}{n^2}$, $b_n:=\frac{1}{n^{3/2}}$. We have that
$$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{\ln (n^2)}{n^{1/2}}=\lim\limits_{n \to \infty}\big(2\cdot \frac{\ln (n)}{n^{1/2}}\big)=0.$$
Therefore, there exists $N$ such that $n>N$ implies $\frac{a_n}{b_n}<1$. Hence, $n> N$ implies $a_n< b_n$, and the comparison test yields convergence.
Aloizio Macedo
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You should make it more clear that you're not using limit comparison test, but rather justifying the usual comparison test. – Stella Biderman Apr 24 '16 at 13:23
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Stolz-Cesaro theorem for $a_n = \log n, b_n = n^2$: $$ \sum_{k=1}^{\infty} \frac{\log (k+1) -\log k}{(k+1)^2 - k^2} \sim \sum_{k=1}^{\infty} \frac{1}{k(2k+1)} \sim \frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^2} $$
The second step is due to Maclaurin series expansion. The last is a famous $\zeta(2)$.
Alex
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