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.

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