Question

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods.
[infinity] n = 1 n2/5n n = 1
lim n→[infinity] an + 1/an =
a. converges
b. diverges

Answer 1

Answer:

A. The series CONVERGES

Step-by-step explanation:

If $\sum a_n$ is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

$\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho$

If $\rho$ < 1, the series converges absolutely

If $\rho > 1$, the series diverges

If $\rho = 1$, the test fails.

Given the series $\sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}$

To test for convergence or divergence using ratio test, we will use the condition above.

$a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}$

$\frac{a_n_+_1}{a_n} = \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}$

$\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}$

aₙ₊₁/aₙ =

$\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|$

note that any constant dividing infinity is equal to zero

$|\frac{1+2/\infty+1/\infty^2}{5}|$

$\frac{1+0+0}{5}\\ = 1/5$

$\rho = 1/5$

Since The limit of the sequence given is less than 1, hence the series converges.