For what r does $\frac{3}{n^{2 r - 3}} - \frac{3}{n}$ $3/(n^(2r - 3))-3/n$ converge or diverge?

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For what r does $\frac{3}{n^{2 r - 3}} - \frac{3}{n}$ $3/(n^(2r - 3))-3/n$ converge or diverge?

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Answer 1 · 2016-05-29T15:08:45

The series sum converges only for $r = 2$ $r = 2$

Explanation:

This series can be decomposed as the sum of two series: $\sum n a_{n}^{1} = \sum n \frac{3}{n^{2 r - 3}}$ $sum_n a_n^1 = sum_n 3/(n^{2r-3}$ and $\sum n a_{n}^{2} = \sum n - \frac{3}{n}$ $sum_n a_n^2 = sum_n -3/n$ so that $\sum n a_{n} = \sum n (a_{n}^{1} + a_{n}^{2})$ $sum_n a_n = sum_n (a_n^1+a_n^2)$ . The series $\sum n a_{n}^{2}$ $sum_n a_n^2$ is allways divergent and $\sum n a_{n}^{1}$ $sum_n a_n^1$ converges or diverges depending on $r$ $r$ . So $\sum n a_{n}$ $sum_n a_n$ is convergent only for $r = 2$ $r = 2$ when $a_{n}^{1} + a_{n}^{2} = 0$ $a_n^1+a_n^2=0$

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For what r does $\frac{3}{n^{2 r - 3}} - \frac{3}{n}$ $3/(n^(2r - 3))-3/n$ converge or diverge?

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Explanation:

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For what r does $\frac{3}{n^{2 r - 3}} - \frac{3}{n}$ $3/(n^(2r - 3))-3/n$ converge or diverge?