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My only guess is to start with $\frac{2a_n+3nb_n}{4n+1}$ = $\frac{2a_n}{4n+1}+\frac{3nb_n}{4n+1}$=$[\frac{2}{4n+1}*(a_n)]$+[$\frac{3n}{4n+1}*(b_n)$] and then I don't really know where to go from there. I'm not even sure if that's going in the right direction.

edit: Okay, so since we know that $\frac{2}{4n+1}$ converges to 0 and $\frac{3n}{4n+1}$ converges to $\frac{3}{4}$, and we know that the product of two convergent sequences is convergent...could we split all of them up and get that $\lim \limits_{n \to \infty}\frac{2}{4n+1}$$\lim \limits_{n \to \infty}a_n$ + $\lim \limits_{n \to \infty}\frac{3n}{4n+1}$$\lim \limits_{n \to \infty}b_n$ = 0*A + $\frac{3}{4}*B$ = $\frac{3B}{4}$? Am I using the limit laws correctly?

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I am going to assume that the limit laws are the algebraic limit theorems or whatever your specific book calls them. Roughly they say that algebra plays nice with limits.

Starting off with $\frac{2}{4n+1}*(a_n) + \frac{3n}{4n+1}*(b_n)$. Let $c_n = \frac{2}{4n+1}$ and $d_n = \frac{3n}{4n+1}$. Notice that $c_n \rightarrow 0$ and $d_n \rightarrow 3/4$. Then by the limit laws $\lim a_n * c_n + b_n * d_n = 0*A + 3/4*B = 3/4 * B$.

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  • $\begingroup$ i actually edited the question with what I thought was how to proceed before I read your response and we ended up with the same answer. Thank you! $\endgroup$ – user21 Feb 28 '17 at 3:08

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