Question

Use the Integral Test to determine whether the series is convergent or divergent

Answer 1

\[\sum_{n=1}^{\infty }n*e^{-12n}\]

Answer 2

please you can use the n-th roots criterion too, since your series is a series of positive terms

Answer 3

what is that?

Answer 4

please note that if the subsequent quantity:
\[\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = L\quad and\;L < 1\]
then the positive terms series
\[\sum\limits_n {{a_n}} \]
is a convergent series

Answer 5

is "e" a square root like that?

Answer 6

Now in your case we have:
\[\begin{gathered}
{a_n} = \frac{n}{{{e^{12n}}}} \hfill \\
\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\frac{n}{{{e^{12n}}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt[n]{n}}}{{{e^{12}}}} = \frac{1}{{{e^{12}}}} < 1 \hfill \\
\end{gathered} \]

Answer 7

oh wow.. thank you!

Answer 8

thank you!