How do you use the limit comparison test on the series $sum_(n=2)^oosqrt(n)/(n-1)$ ?

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Answer 1 · 2015-07-29T16:15:43

The series diverges.

Note first that we can write:

$sum_(n = 2)^oo sqrtn/(n-1)$ as $sum_(n=1)^oo sqrt(n+1)/n$

In the limit, the terms 'act like' $1/sqrtn$ so we'll use the series:

$sum_(n=1)^oo 1/sqrtn$ for the limit comparison test.

$c = lim_(nrarroo) (sqrt(n+1)/n)/(1/sqrtn)$

$= lim_(nrarroo) (sqrt(n+1)sqrtn)/n$

$= lim_(nrarroo) sqrt(n^2+n)/n$

$= lim_(nrarroo) (nsqrt(1+(1/n)))/n$

$= 1$

Since $c$ is positive and finite the two series either both converge or both diverge.

$sum_(n=1)^oo 1/sqrtn$ diverges by comparison with the harmonic series, so

$sum_(n=1)^oo sqrt(n+1)/n$ diverges, as does the original series.

$sum_(n = 2)^oo sqrtn/(n-1)$ diverges.

How do you use the limit comparison test on the series sum_(n=2)^oosqrt(n)/(n-1)sum_(n=2)^oosqrt(n)/(n-1) ?