Question

Prove convergence or divergence for the series:
∑∞ n=0 [(2n+1)/3n+2)]

∑∞ n=0 [(2n-1)/(3n3 + 2n + 4)]

∑∞ n=1 [absolute value of (sin n)] / n2

Answer 1

State test used as well plz

Answer 2

For the first one, use the test for divergence. For a sum to converge we must have that:
\[\lim_{k \rightarrow \infty}a_k=0\]
For this we have:
\[\lim_{n \rightarrow \infty}\frac{2n+1}{3n+2}=2/3 \ne 0 \implies \sum_j \frac{2j+1}{3n+2} D.N.C.\]
Does not converge.

Answer 3

Thanks can you do this for the others as well?

Answer 4

Well for the second one you want to see what it does in the long term, as n gets large it is dominated by:
\[\frac{2}{3n^2}\]
So simply pick a CONVERGENT series that is BIGGER.
Say...
\[\sum_k \frac{1000}{k^2}\]
Which is GREATER than your series for large n. Therefore by limit comparison SINCE 1000/k^2 CONVERGES, then your series converges.
http://www.wolframalpha.com/input/?i=sum+n%3D0+to+infinity+of+(2n-1)%2F(3n%5E3%2B2n%2B4)

Answer 5

Thanks so much and for the third one?

Answer 6

\[\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n^2}\]
Well we know that sin(n) goes between -1 and 1 and the absolute value restricts that to just between 0 and 1, this means that a series such as 3/n^2 would be bigger therefore:
\[\sum_n \frac{|\sin(n)|}{n^2} < \sum_n \frac{3}{n^2} \]
and 3/n^2 converges so by limit comparison the other converges.
http://www.wolframalpha.com/input/?i=sum+from+n+%3D+1+to+infinity+of+%7Csin(n)%7C%2Fn%5E2
It apparently converges to like 1.3ish

Answer 7

Thanks so much!

Answer 8

No problem :P