Question

Does this series converge or diverge? Which method did you use? 1/(n(n+3)) w/ sigma to infinity n=1

Answer 1

1/3 \[\int\limits_{1}^{infi}\] ((n+3) - n)/n(n+3)

Answer 2

now you can solve the integration easily...

Answer 3

if you worked this out, did you get 11/18? and do you mind also telling me if n+1/2n-1 converges? to 1/2?

Answer 4

Comparison method would work here as well.

Answer 5

Ratio would work too actually.

Answer 6

I think.

Answer 7

But comparison seems to be the easiest.

Answer 8

\[
n(n+3)>n(n) \implies \frac{1}{n(n+3)} < \frac{1}{n^2}
\]

Answer 9

Yep that was my guess too.

Answer 10

That converges obviously.

Answer 11

Because the exponent is greater than 1.

Answer 12

i think you can actually compute this one exactly

Answer 13

It has to be an telescoping series though for that.

Answer 14

it does telescope

Answer 15

P-series test.

Answer 16

Ohh yes it does! :P .

Answer 17

Forgot about tpartial fractions.

Answer 18

\[\sum\frac{1}{n(n+3)}=\frac{1}{3}\sum \frac{1}{n}-\frac{1}{n+3}\]

Answer 19

The series in the first question is telescoping so they might be talking about that one?

okay one more, can you help with a method to solve this one? (1/n) - (1/n+2)

Answer 20

For that one I would use the integral test.

Answer 21

And by the looks of it, it's going to diverge.

Answer 22

that one telescopes too

Answer 23

Really? It looks like it's going to diverge to me.

Answer 24

if you are going to use a test for convergence, and not actually add, you have to add them up first