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Would the following series be convergent or divergent.

$\sum\frac{k^k}{3^{k^2}}$

I applied the root test to the following series and got.

$(\frac{k^k}{3^{k^2}})^{1/k}$

I got $\lim$ $k\rightarrow\infty$

$\frac{k}{3^k}$

using Le Hospitals rule I got

$\frac{1}{3^k}$

which is zero

0<1 convergent?

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  • $\begingroup$ Except for a constant in the denominator after applying l'Hopital's rule (which doesn't change the limit), it looks good. $\endgroup$ – Clayton Nov 8 '13 at 18:50
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Correct limit and correct conclusion; you simply have a slight error in the derivative of the denominator, $(3^k)' = (\log 3)3^k$.

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  • $\begingroup$ Oh yes that makes sense $\endgroup$ – Fernando Martinez Nov 8 '13 at 18:58

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