For what r does #n/(n^(r))-3/n^2# converge or diverge?

1 Answer
Cesareo R.
Aug 17, 2016

#r > 2#

Explanation:

Supposing that you are interested in the series sum

#sum_{n=1}^oon/(n^(r))-3/n^2# This series is composed of two terms. The second sum term is convergent. The first series term

#n/n^r = 1/n^ {r-1}# is convergent for #r-1 > 1# or #r > 2#

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