Question

calc-Determine whether the series is convergent or divergent. If convergent, find the sum

Answer 1

using divergence/integral test
\[\sum_{n=1}^{\infty} \frac{ n }{ \sqrt{n^2 + 8} }\]

Answer 2

You could do divergence test first, because it's easier

Answer 3

can someone explain how?

Answer 4

@darkigloo

For the divergence test:

You have to take the limit of the series as it approaches infinity. If it goes to infinity, it diverges. If it goes to 0, the test fails. So for this one:

\[\lim_{n \rightarrow \infty}\frac{ n }{ \sqrt{n^2+8} }\]

This goes to 0 because the bottom grows faster than the top. We would want to say converge, but the divergence test doesn't tell us this, so we must try the integral test.

For the integral test, we take the improper integral of this series. If the integral diverges, the series diverges. If the integral converges, the series converges.

\[\int\limits_{1}^{\infty}\frac{ n }{ \sqrt{n^2+8} }\]

Good ol' u-sub should work

\[u = n^2+8\]\[du=2dx\]

\[\frac{ 1 }{ 2\sqrt{u} }du\]

Pull out the constant and blah blah, you'll eventually get that:

\[[\sqrt{x^2+8}]_{1}^{\infty}\]

For improper, we would take the (limit of infinity - limit of 1). This diverges, so the series diverges