# sequence and series, limits

You need to show us what you've tried and let us know what is confusing you otherwise we can't really help or interact with you. Please read the community rules. A post should look something like this:

details...

# Attempt

details...

As far as your actual problem goes...

$$\lim_{n \to \infty}n(\frac{\cos(\frac{2n+1)}{n^2+n})\sin(\frac{1}{n^2+n})}{\cos(\frac{2n+3}{(n+1)^2+n+1})\sin(\frac{1}{(n+1)^2+n+1})}-1)$$

Since $$n \to \infty$$ we really only need to worry about the dominant terms so we can look at

$$\lim_{n \to \infty}n(\frac{\cos(\frac{1}{n})\sin(\frac{1}{n^2})}{\cos(\frac{1}{n})\sin(\frac{1}{n^2})}-1)$$

and that should make the answer fairly clear.

• @Rhythmlnk - i got answer of 2 when i solve it – Dudley Oct 22 '18 at 23:53
• Not quite. The limit should be 0. Can you tell me why? – Aaron Zolotor Oct 22 '18 at 23:54
• $\sum _{n=1}^{\infty }\:\cos \left(\frac{2n+1}{n^2+n}\right)\sin \left(\frac{1}{n^2+n}\right)$ actually this the problem @Rhythmlnk – Dudley Oct 23 '18 at 0:06
• I don't know what you're asking. Please edit your question so people can understand what you're looking for. – Aaron Zolotor Oct 23 '18 at 2:34

Considering $$a_n=\cos \left(\frac{2n+1}{n^2+n}\right)\sin \left(\frac{1}{n^2+n}\right)$$ Use the classical expansions of cosine and sine and continue with Taylor series to get $$\cos \left(\frac{2n+1}{n^2+n}\right)=1-\frac{2}{n^2}+\frac{2}{n^3}+O\left(\frac{1}{n^4}\right)$$ $$\sin \left(\frac{1}{n^2+n}\right)=\frac{1}{n^2}-\frac{1}{n^3}+O\left(\frac{1}{n^4}\right)$$ making $$a_n=\frac{1}{n^2}-\frac{1}{n^3}+O\left(\frac{1}{n^4}\right)$$ $$\frac{a_n}{a_{n+1}}=\frac{\frac{1}{n^2}-\frac{1}{n^3}+O\left(\frac{1}{n^4}\right) }{\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}+O\left(\frac{1}{n^4}\right) } \approx \frac{(n-1) (n+1)^3}{n^4}=1+\frac 2n+\cdots$$