# Why $\sum_{n=1}^\infty {\ln(n^2) \over n^2 }$ converges? [duplicate]

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.

## marked as duplicate by rtybase, Yanior Weg, Jyrki Lahtonen, Cesareo, user3658307May 11 at 18:43

• First, $\ln(n^2)= 2\ln(n)$. – Omnomnomnom Apr 24 '16 at 13:19
• You can note, for example, that $$\ln(n)/n \leq 1/\sqrt n$$ for sufficiently large $n$ – Omnomnomnom Apr 24 '16 at 13:21

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.

• 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

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.

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 $\zeta(2)$.