11.6: Comparison Test - Mathematics LibreTexts
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/11%3A_Sequences_and_Series/11.06%3A_Comparison_Test
WEBDec 21, 2020 · Solution. The obvious first approach, based on what we know, is the integral test. Unfortunately, we cannot compute the required antiderivative. But looking at the series, it would appear that it must converge, because the terms we are adding are smaller than the terms of a \ (p\)-series, that is, \ [ {1\over n^2\ln n} < {1\over n^2},\]
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