Use divergent test to determine convergence of the following series?

Question

$\infty \sum n = 0 (\frac{3^{n + 1}}{e^{n}})$ $sum_(n=0)^oo(3^(n+1)/e^n)$

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Answer 1 · 2018-03-25T05:47:46

Diverges.

The divergence test, when applied to a series $\sum a_{n}$ $suma_n$ simply entails taking $lim_(n->oo)a_n$ :

If $lim_(n->oo)a_n ne0,$ the series $suma_n$ diverges.

Note, if $lim_(n->oo)a_n=0,$ the series doesn't necessarily converge. It may, but it's not guaranteed.

So,

$lim_(n->oo)3^(n+1)/e^n=lim_(n->oo)(3^n3^1)/e^n=3lim_(n->oo)(3/e)^n=3(oo)=oo$

Thus, the series diverges.