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Use the Ratio Test to determine whether the series is convergent or divergent.

sum_(k=1)^(oo)9ke^(-k)
Identify
a_(k).
Evaluate the following limit.

lim_(k rarr oo)|(a_(k)+1)/(a_(k))|

◻
Since
lim_(k rarr oo)|(a_(k)+1)/(a_(k))|?vv1,- Select--

Use the Ratio Test to determine whether the series is convergent or divergent. $\newline$ $\sum_{k=1}^{\infty} 9 k e^{-k}$ $\newline$ Identify $a_{k}$ . $\newline$ Evaluate the following limit. $\newline$ $\lim _{k \rightarrow \infty}\left|\frac{a_{k}+1}{a_{k}}\right|$ $\newline$ $\square$ $\newline$ Since $\lim _{k \rightarrow \infty}\left|\frac{a_{k}+1}{a_{k}}\right| ? \vee 1,-$ Select--

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Simplify radical expressions with variables II

Full solution

Q. Use the Ratio Test to determine whether the series is convergent or divergent.

\newline

\sum_{k=1}^{\infty} 9 k e^{-k}

\newline

Identify

a_{k}

.

\newline

Evaluate the following limit.

\newline

\lim _{k \rightarrow \infty}\left|\frac{a_{k}+1}{a_{k}}\right|

\newline

\square

\newline

Since

\lim _{k \rightarrow \infty}\left|\frac{a_{k}+1}{a_{k}}\right| ? \vee 1,-

Select--

Identify General Term: Identify $a_{k}$ as the general term of the series, which is $9ke^{-k}$ .
Evaluate Limit: Evaluate the limit $\lim_{k \to \infty}\left|\frac{a_{k}+1}{a_{k}}\right|$ . Start by finding $a_{k}+1$ which is $9(k+1)e^{-(k+1)}$ .
Calculate Ratio: Now, calculate $\frac{a_{k}+1}{a_{k}} = \frac{9(k+1)e^{-(k+1)}}{9ke^{-k}} = \frac{k+1}{k}e^{-1}$ .
Simplify Expression: Simplify $\left(\frac{k+1}{k}\right)e^{-1}$ to $\left(1 + \frac{1}{k}\right)e^{-1}$ .
Take Limit: Take the limit $\lim_{k \to \infty}(1 + \frac{1}{k})e^{-1}$ . As $k$ approaches infinity, $\frac{1}{k}$ approaches $0$ , so the expression simplifies to $e^{-1}$ .
Apply Ratio Test: Since $e^{-1}$ is less than $1$ , the Ratio Test tells us that the series is convergent.

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