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Determine whether the following series is convergent or not, with explanation. $$ \begin{equation} \sum\limits_{n = 1}^\infty {\frac{{{{\left( {2 + \sin n} \right)}^n}}}{{{3^n} \cdot n}}} \end{equation} $$ I guess the above series is disvergent, but I cannot prove it. I have the following assumptions:

  1. The function $\sin n$ has least upper bound one for counting number $n$.

  2. In each 'cycle', there will be a positive number $n_k$ which lets $\sin n_k \to 1$.

Then I just consider all positive numbers $n_k \left(k=1,2,\cdots\right)$ $$ \begin{equation} \sum\limits_{{n_k}} {\frac{{{{\left( {2 + \sin n} \right)}^n}}}{{{3^n} \cdot n}}} \to \sum\limits_{{n_k}} {\frac{1}{n}}, \end{equation} $$ it looks like a harmonic series.

PS.

I used MATLAB to find the maximum of $\sin n$ within a certain range.

n=1:1e4;[m,index]=max(sin(n));

Then, I got $\rm{m}=1.0000$ and $\rm{index}=9929$.

Milan
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Zeta
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  • I would break up the general term with the binomial theorem and then show each sum converges. But i am not sure it works – Milan Aug 25 '19 at 14:25
  • @Milan, while this would work with a finite n, a sum of infinite convergences is not necessarily convergent. – Gabe Aug 25 '19 at 14:52
  • Sin n can only become 1 if n is transcendental? But n is a counting number. – imranfat Aug 25 '19 at 15:02
  • hint: try to look at it as a geometric sequence with a reason smalker than 1 multiplied by $1/n$ – Souames Aug 26 '19 at 00:51
  • @Souames, I know that, however, there is an infinite number of convergence points like a geometric sequence and an infinite number of divergence points like a harmonic series. – Zeta Aug 26 '19 at 01:00
  • The numerator is always less than $3^n$, so each term is (at least a bit) smaller than $1/n$. I'd guess convergent. Try the root test (many powers)... – vonbrand Aug 26 '19 at 01:16
  • I’d guess convergent too, but I definitely don’t know, since I expect that $\forall \epsilon >0, sin(n) > 1-\epsilon$ infinitely often. – Joe Aug 26 '19 at 01:30
  • We must find some kind of precise measure of just how often that happens. – Ivan Neretin Aug 27 '19 at 14:54
  • This property could help https://math.stackexchange.com/q/3333804/503397 – Milan Aug 27 '19 at 15:52
  • Also this https://math.stackexchange.com/q/3337261/503397 – Milan Aug 28 '19 at 18:22
  • It seems that Raab's test, Bertrands's test, Gauss's test all fail, and do not provide a conclusion. https://en.wikipedia.org/wiki/Ratio_test#4._Gauss’s_test – GEdgar Sep 05 '19 at 12:14

1 Answers1

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Not sure if this is ok as an answer, but I can't comment yet. I put it in Wolfram Alpha, which says it doesn't converge. Not sure how it tests that, though.

About Jones' answer, I'm not sure you can claim the strict inequality after taking the limit.

Yannick Neyt
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