Question

doing the integral test for sequences and series

Answer 1

\[= -1/(4(2n+1)^2) from 1 \to \infty \]

Answer 2

i have done the integration just kinda stuck on the step of solving for convergence

Answer 3

\[\sum_{n=1}^\infty -\frac{1}{4(2n+1)^2}~~?\]

Answer 4

no i had \[\sum_{1}^{\infty} \frac{ 1 }{ (2n+1)^3 }\]

Answer 5

n=1*

Answer 6

Okay, well the first thing to do is to show that the sequence is positive and decreasing. It shouldn't be too hard to see that it is.

Now, when you integrate, if the definite integral converges, then so does the series.
\[\int_1^\infty \frac{dx}{(2x+1)^3}=\lim_{b\to\infty}\int_1^b\frac{dx}{(2x+1)^3}=\cdots\]

Answer 7

You said you've computed the integral - what answer did you get?

Answer 8

i got −1/(4(2n+1)2) for my integral then \[\infty \] went to 0 so
0-(-1/36) showed it was convergent?

Answer 9

Yes, if you get a finite number after computing the integral, then the series converges.

And just to check:
\[\lim_{b\to\infty}\int_1^b\frac{dx}{(2x+1)^3}=\frac{1}{2}\lim_{b\to\infty}\int_3^b\frac{du}{u^3}=-\frac{1}{4}\lim_{b\to\infty}\left[\frac{1}{u^2}\right]_3^b=-\frac{1}{4}\left(\lim_{b\to\infty}\frac{1}{b^2}-\frac{1}{9}\right)=\frac{1}{36}\]
Hence convergent.

Answer 10

* 1/36

Answer 11

thank you so much, i was on the right track and was correct just needed the reassurance :D