How do you use basic comparison test to determine whether the given series converges or diverges for #sum n/sqrt(n^2-1)# from n=2 to #n=oo#?

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Answer 1 · 2015-05-15T08:06:47

The series diverges.

Let's begin with an inequality :

#n^2 - 1 <= n^2#

#sqrt(n^2-1) <= sqrt(n^2) = n#, with #n>=1#

#1/(sqrt(n^2-1)) >= 1/n #, with #n >1#

#n/sqrt(n^2-1) >= n/n = 1#

#sum_{n=2}^infty(n/sqrt(n^2-1)) >= sum_{n=2}^infty(1)#

And obviously, #sum_{n=2}^infty(1) = +oo#, so it diverges.

By comparison test, #sum_{n=2}^infty(n/sqrt(n^2-1))# diverges too.