Limit regarding $a_{k}b_{n-k}$

The sequences ${a_n}$ and ${b_n}$ can be represented as $f(n) + 3$ and $g(n) + 5$ respectively where $f(n)$ and $g(n)$ may or may not be equal to each other and $\lim_{n \to \infty} f(n) = 0$ and $\lim_{n \to \infty} g(n) = 0$ . If we substitute these formulas for ${a_n}$ and ${b_{n-k}}$ we get:

$\displaystyle \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}\left(f\left(k\right)+3\right)\left(g\left(n-k\right)+5\right)$

Which if we expand gives us

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\left(f\left(k\right)g\left(n-k\right)+3g\left(n-k\right)+5f\left(k\right)+15\right)$

The $15$ can be easily factored out to become $15n$ . As for the rest of the terms, $\sum_{k=1}^{\infty}f\left(k\right) = \text{constant}$ since $\lim_{n \to \infty}f(n) = 0$ and $\sum_{k=1}^{\infty}g\left(n - k\right) = \text{constant}$ since $\lim_{n \to \infty}g(n) = 0$ and $\sum_{k=1}^{n}g\left(n-k\right)=\sum_{k=0}^{n-1}g\left(k\right)$ . In the case of, if we take $n$ to infinity, $\sum_{k=1}^{n}f\left(k\right)g\left(n-k\right) = f(1)g(n - 1) + f(2)g(n - 2) + ... + f(n- 1)g(1) + f(n)g(0)$ we always have one "large" number countered by a "really small number" (for instance, $f(1)$ is relatively large compared to $g(n - 1)$ which is near zero). As such, the sum should be a constant.

Therefore, we have:

$\displaystyle \sum_{k=1}^{n}\left(f\left(k\right)g\left(n-k\right)+3g\left(n-k\right)+5f\left(k\right)\right) = \text{constant}$

meaning that the limit simplifies to

$\displaystyle \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}(\text{constant} + 15n) = 0 + 15 = \boxed{15}$

Here's a "fakesolve" solution:

Let $\{a_n\} = 3$ for all $n$ , and $\{b_n\} = 5$ for all $n$ . Then it is obvious that $a_k b_{n-k}$ is equal to a constant, $15$ . Because the summation means we are adding $15$ $n$ times, the limit evaluates to $\displaystyle \lim_{n\rightarrow \infty} \dfrac{15n}{n}$ , which simplifies to $\boxed{15}$ .

Obviously, the question asks to prove it for general $\{a_n\}$ and $\{b_n\}$ , but this is a start.