So I have a sequence $\epsilon(n)$ that converge to 0 when $n$ goes to infinity. What can I say about the series for $n \rightarrow \infty$ $$\sum{\frac{\epsilon(n)}{n}}?$$ Thanks for your help.
Convergence of a generic series
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calculus
sequences-and-series
examples-counterexamples
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0Wolfram says $\frac1{n\log n}$ does not converge. – 2017-01-25
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0@RiccardoOrlando in fact all series of the form $\frac{1}{n\left(\log n\right)^p}$ diverge for $p\leq 1$. https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#ln-series – 2017-01-25
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2But why provide infinitely many counterexamples, when one does the trick? :P – 2017-01-25
2 Answers
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Nothing. The series will converge if $\epsilon(n) = 1/n$, and will diverge if $\epsilon(n) = 1 / \log(n)$.
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If $\lim_{n\to \infty}\epsilon(n)=0$, the series could (i) converge absolutely, (ii) diverge, or (iii) converge conditionally.
If $\epsilon(n)=\frac1n$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{1}{n^2}$ is seen to converge absolutely by using, for example, the integral test.
If $\epsilon(n)=\frac{1}{\log(n)}$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{1}{n\log(n)}$ is seen to diverge, by using, again for example, the integral test.
And if $\epsilon(n)=\frac{(-1)^n}{\log(n)}$, then the series $\sum_{n=1}^\infty \frac{\epsilon(n)}{n}=\sum_{n=1}^\infty \frac{(-1)^n}{n\log(n)}$ converges conditionally using Leibniz's test.
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0Please let me know how i can improve my answer. I really want to give you the best answer I can. -Mark – 2017-01-25