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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

1 Answer

5 votes

Answer:

A. The series CONVERGES

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) |(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 (n^2)/(5^n)

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


a_n = (n^2)/(5^n) \\a_n_+_1 = ((n+1)^2)/(5^(n+1))


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


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

aₙ₊₁/aₙ =


\lim_(n \to \infty) |( n^2+2n+1)/(5n^2)| \\\\Dividing\ through\ by \ n^2\\\\\lim_(n \to \infty) |( n^2/n^2+2n/n^2+1/n^2)/(5n^2/n^2)|\\\\\lim_(n \to \infty) |(1+2/n+1/n^2)/(5)|\\\\

note that any constant dividing infinity is equal to zero


|(1+2/\infty+1/\infty^2)/(5)|\\\\


(1+0+0)/(5)\\ = 1/5


\rho = 1/5

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

User Erika
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