Investigate convergence of $\sum_{n=1}^\infty \frac{\ln(n)}{n}$
I applied nth term test and was inconclusive.
I tried ratio test but I don't know how to evaluate the limit. I think it is 1 therefore also inconclusive.
Anyone have any ideas?
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I was thinking by saying this series is greater than the harmonic series but this is untrue for smaller values of n. – user2250537 Apr 28 '16 at 17:50
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1It does not matter that this is untrue for small values of n. What matters is that $\ln(n) > 1$ for sufficiently large $n$. (in this case for $n>1$) – Dark Apr 28 '16 at 17:52
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3For convergence/divergence, the first $2$ terms, or the first $20000$, do not matter. – André Nicolas Apr 28 '16 at 17:52
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1Ok so I could compare it to the harmonic series then? – user2250537 Apr 28 '16 at 17:55
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@Dark $n>2$, actually. – Did Apr 28 '16 at 17:57
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@Did indeed, thanks. – Dark Apr 28 '16 at 17:59
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1Compare it to $\sum_{n=1}^\infty \frac{1}{n}$. – Bonnaduck Apr 28 '16 at 18:00
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Ok great thanks! – user2250537 Apr 28 '16 at 18:00
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2$\ln(n)/n\neq \ln(n/e^n)$. You are misremembering the logarithm rule: we have $\ln(a)-\ln(b)=\ln(a/b)$, not $\ln(a)/\ln(b)=\ln(a/b)$ – Wojowu Apr 28 '16 at 18:08
2 Answers
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Hint 1: $(n > e) \Rightarrow \ln n > 1$
Hint 2: Neglecting finite number of elements from a series doesn't change it's convergence.
Tacet
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Because the sequence defining this series is eventually monotonically decreasing (that is, there exists a point after which the sequence is monotonically decreasing), we can apply the Cauchy condensation test: $$\sum_{n=3}^\infty 2^n \frac{\ln(2^n)}{2^n}=\ln(2)\sum_{n=3}^\infty n= \infty$$
Therefore $\sum_{n=1}^\infty \frac{\ln(n)}{n}=\infty$.
Squirtle
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