sequence and series, limits

Question

Struck with this expression while solving, can you help me solving this

Answer 1

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Problem

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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.

Answer 2

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