Question

Need help telling if this series converges or diverges..

Answer 1

\[\sum_{n=0}^{\infty} n/( n^2+3)\]

Answer 2

It diverges. \[ n^2 +3 \approx n^2
\] near \[ \infty\]
So \[ \frac{n}{n^2 +3}\approx \frac n{n^2} =\frac 1 n
\]\
Hence the series diverges as the harmonic series
\[ \sum_1^\infty \frac 1 n
\]

Answer 3

It says use the nth term test... Is this it?

Answer 4

No it is not. It is called the Limit Comparison Test. See
http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx
Towards the end.