How do you use the ratio test to test the convergence of the series $∑ (n+1)/(3^n)$ from n=1 to infinity?

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Answer 1 · 2018-04-05T21:45:52

By the ratio test, the series converges.

The ratio tests states that a series $sum_n^oo a_n$ converges if $L<1$ and diverges if $L>1$ , where

$L = lim_(n->oo) |a_(n+1)/a_n|$

If the limit is $1$ or doesn't exist, the test is inconclusive.

Since the general term of our series is $a_k = (k+1)/3^k$ , we have:

$L = lim_(n->oo) |((n+1)/3^(n+1))/(n/3^n)|$

Note that, as $n->oo$ , this limit is clearly positive therefore the absolute value is not needed.

$L = lim_(n->oo) (n+1)/3^(n+1) * 3^n/n = lim_(n->oo)3^n/(3^n*3) * (n+1)/n$

$L = lim_(n->oo) 1/3 * (1 + 1/n)$

Since $n$ approaches infinity, $1/n$ must approach $0$ .

$L = 1/3 * (1+0) = 1/3$

$L<1$ , which means that the series $color(red)("converges by the ratio test")$ .

How do you use the ratio test to test the convergence of the series ∑ (n+1)/(3^n)∑ (n+1)/(3^n) from n=1 to infinity?