Evaluating a limit proof (from riemann sums)

Question

https://imgur.com/a/F0HLTpg

I was able to prove part a) using upper and lower Riemann sums but I have no clue where to go for part b). How do I prove this?

Answer 1

Though this question is old, I'll answer it out of the desire to help many people.

(a)

Note that the function x → x^α is monotonically increasing for all x > 0. Consequently, the minimum of this function over an interval of the form [i, i + 1] is found at x = i, and the maximum at x = i + 1 for all positive i.

By term-by-term summation, the desired inequality is established.

(b)

Note that the inequality above applies for all n. Apply it inductively to n + 1 and get the inequality:

$\int_0^n x^\alpha dx \le 1^\alpha + \dots + n^\alpha \le \int_0^{n+1} x^\alpha dx$

I don't know how Wyzant handles TeX expressions; I'll give a textual approximation in case it's not handled well:

∫₀ⁿ x^α dx ≤ 1^α + ⋅⋅⋅ + n^α ≤ ∫₀ⁿ⁺¹ x^α dx

Then, apply the power rule to the integrals like so:

n^α+1/(α+1) ≤ 1^α + ⋅⋅⋅ + n^α ≤ (n+1)^α+1/(α+1)

Divide all sides by n^α+1:

1/(α+1) ≤ (1^α + ⋅⋅⋅ + n^α)/(n^α+1) ≤ (n+1)^α+1/(α+1)/(n^α+1)

Interpret the compound fraction to the right like this: 1/2/3 = 1/6, not 3/2.

Now, apply the squeeze theorem, which proves the statement.