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

Question

4464 views around the world

You can reuse this answer
Creative Commons License

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.