<s>[INST] What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
 [/INST] One way is by using the Euler-Maclaurin summation formula, something you strangely don't find in most standard calculus books.
http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula
Let $f(x) = x^4$.
Then
$$f'(x) = 4x^3, f'''(x) = 24 x, f^{(5)}(x) = f^{(7)}(x) = ... = 0 $$
And according to the formula,
$$ \sum_{k=1}^{n} k^{4} = \sum_{k=0}^{n} k^{4} = \int_{0}^{n} x^{4} \ dx - B_{1} \left( f(n) - f(0) \right) + \frac{B_{2}}{2!} \left(f'(n) - f'(0) \right) + \frac{B_{4}}{4!} \left(f'''(n) - f'''(0) \right)$$
where $B_{i}$ is a Bernoulli number.
http://en.wikipedia.org/wiki/Bernoulli_number
So
$$\sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{1}{2} (n^{4}) + \frac{1}{2} \left( \frac{1}{6} \right) (4n^{3}) + \frac{1}{24} \left(-\frac{1}{30}\right)(24n) $$
$$ = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}$$
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