does the series $\sum \frac{8n}{(\log n)^n}$ converge or diverge? I've 'tried' using the root test but I just can't seem to get my head around the log in the denominator and what rules to apply because of it. Can anyone help me out here please? Thanks.
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what exactly is the denominator? – Thomas Apr 11 '14 at 11:48
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(Lnn)^n ... my bad i wrote it wrong, going to try edit the orgiginal now... – Lewis Apr 11 '14 at 11:51
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I've tried to edit the question.. it is now correct or did you mean something else? – Ant Apr 11 '14 at 11:53
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log composed n times or log power n ? – T_O Apr 11 '14 at 11:55
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$\sum \frac{8n}{(logn)^n}$ <--- – Lewis Apr 11 '14 at 11:56
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$2<\log (n)$ from some point on. – Ian Mateus Jul 13 '15 at 22:26
2 Answers
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If $e < a < b$, then $a^b > b^a $. Therefore, for $n > e^e$, since $e < \ln n < n$, $(\ln n)^n > n^{\ln n} $ so $\frac1{(\ln n)^n} < \frac1{n^{\ln n}} $.
Therefore $\frac{n}{(\ln n)^n} < \frac{n}{n^{\ln n}} = \frac{1}{n^{\ln n-1}} $.
Since $\sum \frac{1}{n^{\ln n-1}} $ converges (since $\ln n - 1 > 1$ for $n > e^2$), so does $\sum \frac{n}{(\ln n)^n} $.
marty cohen
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the series converges; the root test yields
$$\lim \frac{\sqrt[n]{8n}}{\log n} = \lim \frac{1}{\log n} = 0 $$
Since the limit is $< 1$ the series converges.
Ant
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