# Math Help - Converge + Diverge

1. ## Converge + Diverge

I have done four problems and I just want to make sure that I am doing them right before I continue.

I have to test for convergence. E= Riemann Sum :P
First problem.

infinite
E n^2/(2n^3+1)
n=1

For this i used Divergence test. Divided all terms by n^3. Ended up with 0/2. This shows that it will converge because it was 0.
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2nd problem.

Infinite
E (-1)^n (2/sqrt(n))
n=1

Bn= 2/sqrt(n) Lim n-> infinity 2/sqrt(n) = 0
Bn = 2/sqrt(n) > 2/sqrt(n+1) = B(n+1)
This shows that it converges
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3rd Problem.

Infinity
E n/2n+3
n=1

Used divergence test again.

divided all 3 terms by n
got 1/2
since not 0 it diverges

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4th problem

Infinity
E (-1)^n+1/((n^2))+3)
n=1

I did the same thing as number 2 and got the same answer.
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If there is a better way to do these, please let me know and I will redo the problem and repost it. Thank you.

2. Originally Posted by PensFan10
I have to test for convergence. E= Riemann Sum (These are not necessarily Riemann sums, they are just infinite series FYI) :P
First problem.
infinite
E n^2/(2n^3+1)
n=1

For this i used Divergence test. Divided all terms by n^3. Ended up with 0/2. This shows that it will converge because it was 0.
This is incorrect, this is one of the most commonly encountered errors. If $\sum{a_n}$ is an infinite series and if $\lim_{n\to\infty}a_n\ne{0}$ then $\sum{a_n}$ diverges. It says nothing of whethere it will converge/diverge if the limit is zero. For this one try comparing your series to $\sum_{n=1}^{\infty}\frac{1}{n}$

2nd problem.

Infinite
E (-1)^n (2/sqrt(n))
n=1

Bn= 2/sqrt(n) Lim n-> infinity 2/sqrt(n) = 0
Bn = 2/sqrt(n) > 2/sqrt(n+1) = B(n+1)
This shows that it converges
This is correct as long as you make sure it is known that you are using Leibniz's Criterion (Alternating series test)

3rd Problem.

Infinity
E n/2n+3
n=1

Used divergence test again.

divided all 3 terms by n
got 1/2
since not 0 it diverges
This is a correct application.

4th problem

Infinity
E (-1)^n+1/((n^2))+3)
n=1

I did the same thing as number 2 and got the same answer.
Yes since $\forall{n}\in\mathbb{N}~a_{n+1}\leqslant{a_n}$ and $\lim_{n\to\infty}a_n=0$ this series does converge by Leibniz's Criterion. But note that this also coverges absolutely.

3. Thanks. For The 4th problem, how can you tell that it converges absolutely? Also, is it possible to use the ratio test on these because when I tried, i got n/n for the first problem. Then for the second one, i got something strange. I figured I canceled wrong or you can't use it on those ones.

Would it be possible if you were to show me the ratio test and the cancellations on the first two?

4. Originally Posted by PensFan10
Thanks. For The 4th problem, how can you tell that it converges absolutely? Also, is it possible to use the ratio test on these because when I tried, i got n/n for the first problem. Then for the second one, i got something strange. I figured I canceled wrong or you can't use it on those ones.

Would it be possible if you were to show me the ratio test and the cancellations on the first two?
$\sum_{n=0}^{\infty}\left|\frac{(-1)^n}{n^2+3}\right|=\sum_{n=0}^{\infty}\frac{1}{n^ 2+3}\leqslant\sum_{n=0}^{\infty}\frac{1}{n^2}$ the last one converges by any number of tests, including integral test.

$\lim_{n\to\infty}\left|\frac{(n+1)^2}{2(n+1)^3+1}\ cdot\frac{2n^3+1}{n^2}\right|\to{1}\therefore\text { inconclusive}$

$\lim_{n\to\infty}\left|\frac{(-1)^{n+1}}{2\sqrt{n+1}}\cdot\frac{2\sqrt{n}}{(-1)^n}\right|\to{1}\therefore\text{ inconclusive}$

5. ## Awesome

thats good that the ratio test doesnt work because that means i didnt do anything wrong. i ended up with the same thing you did. thank you so much