Question

Need help with telling if these series converges or diverges using a comparison test.

Answer 1

\[\sum_{1}^{\infty} n/(\sqrt[]{n+3})\]

Answer 2

Honestly just not sure what comparison test I should use, and what series should be used to compare

Answer 3

It looks like you could compare it to series\[\sum_1^\infty {n \over \sqrt{n^2}} = \sum_1^\infty {n \over n} \]Since \(\sqrt{n+3} \leq \sqrt{n^2}\) For \(n>2\)

Answer 4

Just compare it to 1+1+1+...

Answer 5

Since \(\sqrt{n+3} \leq \sqrt{n^2}\), \[{1 \over \sqrt{n+3}} \geq {1 \over \sqrt{n^2}}\]Hence, your original sum is greater than the sum \(1+1+1+...\)

Answer 6

that makes sense I guess then... let me try it

Answer 7

So the n's not matter on top since they are the same?can just write it as 1?

Answer 8

\[\sum_1^\infty {n \over n} =\sum_1^\infty 1\]By cancelling the \(n\)'s.

Answer 9

So would I use limit, or direct?

Answer 10

Limit right?

Answer 11

To sum everything up, \[\sum_{1}^{\infty} {n \over \sqrt[]{n+3} }\geq \sum_1^\infty {n \over \sqrt{n^2}} = \sum_1^\infty {n \over n}=\sum_1^\infty 1\]Thus, your series diverges.

Answer 12

Oh yeah.. smaller diverges, so bigger will diverge because of direct comparison! Thank you. I kept thinking I needed to use 1/n or 1/n^2

Answer 13

you're welcome