2022-01-22

I want to prove that
$\sum _{x=1}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}}\le 1$

basgrwthej

Expert

Note that
$\int \frac{\mathrm{ln}x}{{x}^{2}}dx=-\frac{1+\mathrm{ln}x}{x}+C$
Since $\frac{\mathrm{ln}x}{{x}^{2}}$ is decreasing for $x>2$, we have
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{ln}2}{4}+\frac{\mathrm{ln}3}{9}+\sum _{n=4}^{\mathrm{\infty }}\frac{\mathrm{ln}n}{{n}^{2}}$
$<\frac{\mathrm{ln}2}{4}+\frac{\mathrm{ln}3}{9}+{\int }_{3}^{\mathrm{\infty }}\frac{\mathrm{ln}x}{{x}^{2}}dx$
$\approx 0.29535+0.69954$
$=0.99489$

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