can someone verify if i'm right about these series

**Cato** · October 10th 2008, 05:34 AM

I have to find whether these series converge.

$\sum {\frac {n+1} {n^2 + 1}}$ by using the limit comparison test, and using $\sum {b_n}$ = $\frac {1} {n}$ i found that it diverges.

As for this example: $\sum {\frac {n!} {n^n}}$ i used the ratio test and found it diverges also.

Can someone check if i am correct, and also help me with this one
$\sum {\frac {1} {2^n - 1 + cos^2(n^3)}}$

**ThePerfectHacker** · October 10th 2008, 05:44 AM

Originally Posted by Cato

I have to find whether these series converge.

$\sum {\frac {n+1} {n^2 + 1}}$ by using the limit comparison test, and using $\sum {b_n}$ = $\frac {1} {n}$ i found that it diverges.

This diverges so you are correct.

As for this example: $\sum {\frac {n!} {n^n}}$ i used the ratio test and found it diverges also.

This is wrong. The ratio that you should get is $\left( \frac{n}{n+1} \right)^n$ . The limit of this is $1/e < 1$ . Therefore it converges.

Can someone check if i am correct, and also help me with this one
$\sum {\frac {1} {2^n - 1 + cos^2(n^3)}}$

$\frac{1}{2^n - 1 + \cos^2(n^3)} \leq \frac{1}{2^n - 1}$

**Cato** · October 10th 2008, 06:01 AM

Thank you, so $\frac{1}{2^n - 1}$ would then diverge when n tends to infinity?

**watchmath** · October 10th 2008, 06:31 AM

Originally Posted by Cato

I have to find whether these series converge.

$\sum {\frac {n+1} {n^2 + 1}}$ by using the limit comparison test, and using $\sum {b_n}$ = $\frac {1} {n}$ i found that it diverges.

As for this example: $\sum {\frac {n!} {n^n}}$ i used the ratio test and found it diverges also.

Can someone check if i am correct, and also help me with this one
$\sum {\frac {1} {2^n - 1 + cos^2(n^3)}}$

You got it right for the first.
For the last one, use the fact that $\cos x\leq 1$ . If i give you a little bit more hint it will be obvious. So try first and let see if you get it or not

Edit: sorry I don't know if someone already give a reply (the ad always deceives me as if there is no reply to the post).
I think it is also bound by 1/2^n which is converges (geometric series).

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