<s>[INST] Calculate $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}$ Is my solution correct?
By Faulhaber's formula, $1^4 + 2^4 + \ldots + n^4 = \frac{6n^5+15n^4+10n^3-n}{30} $.
$$\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = \frac{1}{n^5} \left[\sum_{k=1}^{2n} k^4 - \sum_{k=1}^{n} (2k)^4 \right] = \frac{1}{n^5} \left[\frac{6\cdot(2n)^5 + o(n^5) - 2^4 \cdot 6 \cdot n^5 - o(n^5)}{30} \right] = \frac{6\cdot2^5 - 2^4 \cdot 6}{30} + o(1) = \frac{16}{5} + o(1)$$
So, $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = 3.2$
 [/INST] It is correct, of course, but it would've perhaps been much simpler if one were to just recognize the expression as the Riemann sum of $\displaystyle\int_0^1(2x)^4dx$.
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