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May 04, 2026

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📝 Recent Posts

This is a Riemann sum. Split the sum:
a_n = \sum_{k=1}^{n} \frac{k}{n} - \frac{1}{3} \sum_{k=1}^{n} \left( \...
= \frac{1}{n} \sum_{k=1}^{n} k - \frac{1}{3n^3} \sum_{k=1}^{n} k^3
Use formulas:
$ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} $
$ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 $
a_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} - \frac{1}{3n^3} \cdot \left(...
= \frac{n+1}{2} - \frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4}
Simplify the second term:
\frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4} = \frac{(n+1)^Question: A hi...

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