I am doing a self study of real analysis and this comes up and I can't find any problems like it:
$$\lim _{n\rightarrow \infty }\sin\biggl(\frac{1}{n}\biggr)\cdot \Biggl(\sec^2\biggl(\frac{1}{3n}\biggr)+\sec^2\biggl(\frac{4}{3n}\biggr)+\cdots +\sec^2\biggl(\frac{3n-5}{3n}\biggr)+\sec^2\biggl(\frac{3n-2}{3n}\biggr)\Biggr)$$
As much as I have been able to do is possibly change it to
$$\lim _{n\rightarrow \infty }\sin\biggl(\frac{1}{n}\biggr)\cdot \sum_{k=1}^{n}\sec^2\biggl(\frac{3k-2}{3n}\biggr)$$
Any help is great! Thanks!
First:
$$\sin(1/n) \sum_{k=1}^n \sec^2 \left ( \frac{3k-2}{3n} \right ) = \frac{1}{n} \sum_{k=1}^n \sec^2 \left ( \frac{3k-2}{3n} \right ) + \left ( \sin(1/n) - \frac{1}{n} \right ) \sum_{k=1}^n \sec^2 \left ( \frac{3k-2}{3n} \right ).$$
That first term is an actual Riemann sum; you are evaluating $\sec^2$ at certain points inside the partition $0,1/n,\dots,1$. So the limit of that is $\int_0^1 \sec^2(x) dx$, which you can calculate.
In the second term, you have that all the terms in the sum are less than $\sec^2(1)$, so the whole sum is less than $n \sec^2(1)$. So that whole term is less than $\sec^2(1)(1-n\sin(1/n))$ in magnitude. That goes to zero because $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$.