How do you use the Root Test on the series $\infty \sum n = 1 {(\frac{5 n - 3 n^{3}}{7 n^{3} + 2})}^{n}$ $sum_(n=1)^oo((5n-3n^3)/(7n^3+2))^n$ ?

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Answer 1 · 2014-10-02T00:04:01

Let

$a_{n} = {(\frac{5 n - 3 n^{3}}{7 n^{3} + 2})}^{n}$ $a_n=({5n-3n^3}/{7n^3+2})^n$ .

By Root Test,

$lim_{n to infty}rootn{|a_n|}=lim_{n to infty}rootn{|({5n-3n^3}/{7n^3+2})^n|}$

by cancelling the nth-root and the nth-power,

$=lim_{n to infty}|{5n-3n^3}/{7n^3+2}|$

by dividing the numerator and the denominator by $n^3$ ,

$=lim_{n to infty}|{5/n^2-3}/{7-2/n^3}|=|{0-3}/{7-0}|=3/7<1$

Hence, the series converges absolutely.

How do you use the Root Test on the series sum_(n=1)^oo((5n-3n^3)/(7n^3+2))^n∞∑n=1(5n−3n37n3+2)nsum_(n=1)^oo((5n-3n^3)/(7n^3+2))^n ?