Which tests to use for convergence?

Oct 2009
31
2
Hey all i was wondering what tests to use for these series, as some of them i have no idea how to tackle, hopefully you can help me out :).



Pointing me in the right direction will help me greatly for my exam tomorrow!
Thanks people.

I had no idea how to attempt any of them, and my lecture notes aint helping me too. For the sin ones, would i use a comparison with 1/n? because its sin (n) ?
 
Oct 2009
4,261
1,836
Hey all i was wondering what tests to use for these series, as some of them i have no idea how to tackle, hopefully you can help me out :).



Pointing me in the right direction will help me greatly for my exam tomorrow!
Thanks people.

I had no idea how to attempt any of them, and my lecture notes aint helping me too. For the sin ones, would i use a comparison with 1/n? because its sin (n) ?

1) \(\displaystyle \left|\frac{n-\sin^2n}{n^3}\right|\leq \frac{n+1}{n^3}\leq 2\frac{1}{n^2}\) ...comparison test for absolute convergence

2) \(\displaystyle \frac{1}{2^n+1}\leq \left(\frac{1}{2}\right)^n\) ...comparison test

3) \(\displaystyle \lim_{n\to\infty}\frac{n}{\ln n} \neq 0\) ... in fact, \(\displaystyle \frac{n}{\ln n}\xrightarrow [n\to\infty]{}\infty\)

4) \(\displaystyle \frac{n+1/n}{n!+n^3}\leq \frac{2n}{n!}=2\frac{1}{(n-1)!}\) , and then use the ratio test to show this last series converges.

Tonio
 
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Oct 2009
31
2
1) \(\displaystyle \left|\frac{n-\sin^2n}{n^3}\right|\leq \frac{n+1}{n^3}\leq 2\frac{1}{n^2}\) ...comparison test for absolute convergence

2) \(\displaystyle \frac{1}{2^n+1}\leq \left(\frac{1}{2}\right)^n\) ...comparison test

3) \(\displaystyle \lim_{n\to\infty}\frac{n}{\ln n} \neq 0\) ... in fact, \(\displaystyle \frac{n}{\ln n}\xrightarrow [n\to\infty]{}\infty\)

4) \(\displaystyle \frac{n+1/n}{n!+n^3}\leq \frac{2n}{n!}=2\frac{1}{(n-1)!}\) , and then use the ratio test to show this last series converges.

Tonio
Thank you very much! +rep'd