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")#.