Does the series converge?
$$\lim_{n \to \infty}\sum_1^n \dfrac{(\log k)^4}{ k^2}$$
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1Dear Robinson, what have you tried? – Inceptio May 11 '13 at 13:18
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Have you tried any convergence tests? – May 11 '13 at 13:18
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4Do you know for any $\alpha>0$, one has $\log n<n^\alpha$ for sufficiently large $n$? – David Mitra May 11 '13 at 13:21
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1https://en.wikipedia.org/wiki/Integral_test_for_convergence – Najib Idrissi May 11 '13 at 13:28
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1See here for the general theorem that you can use. – Mhenni Benghorbal May 11 '13 at 14:05
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hints: (i) $\ln k$ grows more slowly than any power of $k$ as $k \to \infty$, (ii) $\sum_{n=1}^\infty \frac{1}{n^p}$ converges for any $p > 1$ by the Integral Test. – Stefan Smith May 11 '13 at 15:05
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Notice that $\int_1^{\infty}\frac{\ln(x)^4}{x^2}dx = \frac{-24-24 \ln(x) -12 \ln(x)^2-4 \ln(x)^3 -\ln(x)^4}{x}|_1^{\infty}=24$. Since this integral converges your series converges by the integral test.
Wintermute
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