Backslide introduced in 9, persisting through 11.2.
Consider the series $\sum_{n=1}^\infty\sin\frac{50}{n^2}$. The terms are eventually positive.
ListPlot[Table[Sin[50/n^2], {n, 1, 40}]]
Which produces this image.
Therefore, I was able to use the Limit Comparison Test.
a[n] := Sin[50/n^2]
b[n] := 50/n^2
Limit[a[n]/b[n], n -> ∞]
Which produced an answer of 1. Because this limit is strictly positive and $\sum_{n=1}^\infty\frac{50}{n^2}$ is convergent ($p$-series with $p=2>1$), by the Limit Comparison Test, the series $\sum_{n=1}^\infty\sin\frac{50}{n^2}$ is also convergent. However,
SumConvergence[Sin[50/n^2], n]
does not produce a "True" response. I also tried:
Assuming[n >= 10, SumConvergence[Sin[50/n^2], n]]
Which was also unsuccessful. Is this something I should report?
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