# sequence and series, limits

## 2 Answers

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# Attempt

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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 $$a_n=\frac{1}{n^2}-\frac{1}{n^3}+O\left(\frac{1}{n^4}\right)$$ $$\frac{a_n}{a_{n+1}}=\frac{\frac{1}{n^2}-\frac{1}{n^3}+O\left(\frac{1}{n^4}\right) }{\frac{1}{(n+1)^2}-\frac{1}{(n+1)^3}+O\left(\frac{1}{n^4}\right) } \approx \frac{(n-1) (n+1)^3}{n^4}=1+\frac 2n+\cdots$$