Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 37

Divergent

Work Step by Step

$\Sigma^{\infty}_{n=1} \ln (\frac{n^{2}+1}{2n^{2}+1})$ $a_{n}=\ln (\frac{n^{2}+1}{2n^{2}+1})$ $\lim\limits_{n \to \infty} a_{n} = \lim\limits_{n \to \infty} \ln (\frac{n^{2}+1}{2n^{2}+1})$ $=\ln (\lim\limits_{n \to \infty}\frac{n^{2}+1}{2n^{2}+1})$ $=\ln (\lim\limits_{n \to \infty}\frac{1+\frac{1}{n^{2}}}{2+\frac{1}{n^{2}}})$ $= \ln (\frac{1+0}{2+0})$, since $\lim\limits_{n \to \infty} \frac{1}{n^{2}} = 0$ $= \ln (\frac{1}{2})$ $= \ln 0.5$ $\approx -0.6931$ $\ne 0$ Since $\lim\limits_{n \to \infty} a_{n} \ne 0$, then the series is divergent.

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