I am testing the convergence of $$\sum_{n=2}^{\infty}(-1)^{\left\lfloor {n^3+n+1 \over 3n^2-1} \right\rfloor}{\ln{n} \over n}$$ Dirichlet's test seems appropriate here as for sufficiently large $n$s ${\ln{n} \over n} \to 0$ monotonically. I can also numerically conclude that the sign of $(-1)^{\left\lfloor {n^3+n+1 \over 3n^2-1} \right\rfloor}$ for $n \ge 6$ alternates in batches of 3 terms, which would constrain the partial sum of $(-1)^{\left\lfloor {n^3+n+1 \over 3n^2-1} \right\rfloor}$. This is enough to apply Dirichlet's test but I would prefer to show that $(-1)^{\left\lfloor {n^3+n+1 \over 3n^2-1} \right\rfloor}$ is constrained analytically. Any hints?
Convergence of $\sum_{n=2}^{\infty}(-1)^{\left\lfloor {n^3+n+1 \over 3n^2-1} \right\rfloor}{\ln{n} \over n}$
2
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real-analysis
sequences-and-series
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0Well this is an alternating series, so you can also use the alternating test to show that it is absolutely convergent. – 2017-01-12
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0In the alternating series test, the sign changes from one term to another. In this example we have 3 terms with the same sign followed by 3 terms with a different sign. Is there a straightforward way to adapt the usual test to this case? – 2017-01-12
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0Well I dont know if it makes a difference .. but if you club the 3 terms of the same sign and form 2 series of opposite signs, then you get 2 series both converging to 0. So i think the test can be done. – 2017-01-12
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0Just group the sets of three terms into one term each, and verify the conditions of the alternating series test. – 2017-01-12
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0@Vik78 You need just a little bit more than that (yet not much). Grouping comes with some caveats -- it'd be easy to say that $\sum_n (-1)^n$ converges otherwise. – 2017-01-12
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0If you group adjacent terms that have the same sign I don't think there's any problem. The point is that the $n$th partial sum of the original series is asymptotically within $\epsilon$ of the $\frac{n}{3}$rd partial sum of the regrouped series, so if the regrouped series converges the original one does as well. In the example of $\sum_n (-1)^n$, if you regroup adjacent pairs of terms you do not get the same nice behavior. – 2017-01-12
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0You need to group a constant number of adjacent terms, and also have that the general term goes to $0$ (which is not the case for $(-1)^n$). As I say, it is not complicated -- but it needs to be said. – 2017-01-12