[College Calc 01-02. Limits of Sequences] #03

Let $a_n$ be the expression inside the limit. Note that

$\left(\frac{10}{3}\right)^2=\frac{100}{9}>8=2^3>\left(\frac{7}{4}\right)^3;\quad |\sin^n n\,|\leq 1 < \frac{1}{2}\left(\frac{10}{3}\right)^{2n}.$

Then we may write

$\frac{1}{2}\left(\frac{10}{3}\right)^{2n} \leq\left(\frac{10}{3}\right)^{2n} + \left(\frac{7}{4}\right)^{3n}+\sin^n n\leq \frac{5}{2}\left(\frac{10}{3}\right)^{2n} \\ \, \\ \frac{100}{9}\left(\frac{1}{2}\right)^{1/n}\leq a_n\leq \frac{100}{9}\left(\frac{5}{2}\right)^{1/n}.$

Note that the $1/n$ in the exponent approaches zero -- $\displaystyle \lim_{n\to\infty}\left(\frac{1}{2}\right)^{1/n} = \lim_{n\to\infty}\left(\frac{5}{2}\right)^{1/n} = 1.$

Thus, due to the squeeze theorem , the three expressions all converge to $\dfrac{100}{9},$ so that $p+q=100+9=\boxed{109}.$

First note that:

$\lim_{n \to \infty} \sqrt[n]{\left(\frac {10}3\right)^{2n} + \left(\frac 74\right)^{3n} - 1} < \lim_{n \to \infty} \sqrt[n]{\left(\frac {10}3\right)^{2n} + \left(\frac 74\right)^{3n} + \sin^n n} < \lim_{n \to \infty} \sqrt[n]{\left(\frac {10}3\right)^{2n} + \left(\frac 74\right)^{3n} + 1}$

Now consider

$\begin{aligned} L & = \lim_{n \to \infty} \sqrt[n]{\left(\frac {10}3\right)^{2n} + \left(\frac 74\right)^{3n} \pm 1} \\ & = \lim_{n \to \infty} \sqrt[n]{\left(\frac {100}9\right)^n + \left(\frac {343}{64}\right)^n \pm 1} \\ & = \lim_{n \to \infty} \frac {100}9\sqrt[n]{1 + \left(\frac {3087}{6400}\right)^n \pm \left(\frac 9{100}\right)^n} \\ & = \frac {100}9 \end{aligned}$

So by squeeze theorem we have $\displaystyle \lim_{n \to \infty} \sqrt[n]{\left(\frac {10}3\right)^{2n} + \left(\frac 74\right)^{3n} + \sin^n n} = \frac {100}9$ and $p+q = 100+9 = \boxed{109}$ .

Here's an overkill approach:

Let's first set all the powers as $n$ only. That is, we want to evaluate $\lim_{n\to\infty}\sqrt[{\small n}]{\left(\frac{100}{9}\right)^{n}+\left(\frac{343}{64}\right)^{n}+\sin^n n}.$

Let us invoke Theorem 3 of Stolz–Cesàro theorem :

If the sequence $(x_n /x_{n-1} )_{n=1}^\infty$ has a limit, then $\displaystyle \lim_{n\to\infty} \sqrt[n]{x_n} = \lim_{n\to\infty} \frac{x_n}{x_{n-1}}$ .

In this case, $x_n := (100/9)^n + (343/64)^n + (\sin n)^n$ . Then, $\begin{array} { r c l } \displaystyle \lim_{n\to\infty} \dfrac{x_n}{x_{n-1}} &=& \dfrac{(100/9)^{n} + (343/64)^n + (\sin n)^n}{(100/9)^{n-1} + (343/64)^{n-1} + (\sin n)^{n-1}} \\ \phantom0 \\ &=& \displaystyle \lim_{n\to\infty}\dfrac{(100/9)^{n} + (343/64)^n + (\sin n)^n}{(100/9)^{n-1} + (343/64)^{n-1} + (\sin n)^{n-1}} \div \dfrac{(100/9)^n}{(100/9)^n}\\ \phantom0 \\ &=& \displaystyle \lim_{n\to\infty} \dfrac{1 + (343/64 \div 100/9)^n + (\sin n \div 100/9)^n}{(100/9)^{-1} + (343/64)^{-1} \cdot (343/64 \div 100/9)^n + (\sin n)^{-1} \cdot (\sin n \div 100/9)^n } \\ \phantom0 \\ &=& \dfrac{1 + 0 + 0}{9/100 + 0 + 0} = \dfrac{100}9. \\ \end{array}$ Hence, the limit in question is also equal to $\frac {100}9$ . The answer is $100 + 9 = \boxed{109} .$