How do you use the Ratio Test on the series $\infty \sum n = 1 \frac{9^{n}}{n}$ $sum_(n=1)^oo9^n/n$ ?

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Answer 1 · 2014-10-05T22:26:08

Let

$a_{n} = \frac{9^{n}}{n}$ $a_n={9^n}/{n}$ . $\Rightarrow a_{n + 1} = \frac{9^{n + 1}}{n + 1}$ $Rightarrow a_{n+1}={9^{n+1}}/{n+1}$

By Ratio Test,

$lim_{n to infty}|{a_{n+1}}/{a_n}|=lim_{n to infty}|{{9^{n+1}}/{n+1}}/{{9^n}/{n}}|$

by cancelling out common factors,

$=lim_{n to infty}|{9n}/{n+1}|$

by pulling 9 out of the limit and dividing the numerator and denominator by $n$ ,

$=9lim_{n to infty}|{1}/{1+1/n}|=9cdot 1/{1+0}=9 ge 1$

Hence, the series diverges.

I hope that this was helpful.

How do you use the Ratio Test on the series sum_(n=1)^oo9^n/n∞∑n=19nnsum_(n=1)^oo9^n/n ?