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

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Answer 1 · 2015-11-10T00:55:37

For $r \geq \frac{3}{2}$ $r>=3/2$ and only for them.

Let $α$ $alpha$ be a real number.
The sequence $(1/n^alpha)_(n \in NN^*)$ converges if and only if $alpha >= 0$ .

Of course, $(1/n^alpha)_(n \in NN^*)$ converges iff $(3/n^alpha)_(n \in NN^*)$ converges...

So, here, the sequence converges iff $2r-3>=0$ iff $r>=3/2$ .

For what r does 3/(n^(2r - 3))3n2r−33/(n^(2r - 3)) converge or diverge?