Simple Stirling: Proving Increasing Continuous Function on [1, inf)

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SUMMARY

The discussion focuses on proving that the function defined by the inequality f(1) + f(2) + ... + f(n-1) ≤ ∫_{0}^{n} f(x) dx ≤ f(2) + f(3) + ... + f(n) is increasing and continuous on the interval [1, ∞). The user proposes using the trapezium rule for integral approximation but encounters difficulties due to the nature of f''(x). They also consider left and right Riemann sums as potential solutions to establish the inequality and clarify the concept of increasing functions in this context.

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  • Understanding of integral calculus, specifically Riemann sums
  • Knowledge of properties of increasing functions
  • Familiarity with the trapezium rule for numerical integration
  • Basic concepts of continuity in mathematical functions
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  • Learn about Riemann sums and their applications in approximating integrals
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Homework Statement



[tex]f(1)+f(2)...+f(n-1) =< \int_{0}^{n} f(x) dx =< f(2) +f(3)+...+f(n)[/tex]

is increasing and continuous on [1, inf)
I'm meat to prove the above, the idea I had was to use the trapezium rule to get an approximation of the integral, but since f''(x) can be either negative or positive I'm stuck as to how to do that. Also I'm in a first year course and I haven't yet learned about Bernoulli numbers or the like, which is what came up when I was looking around for this problem.

Edit: Would using left and right Riemann sums solve this? I can get the inequality but does increasing in this case mean non-decreasing?
 
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Are you sure that the integral isn't from 1 to n?

I'd be inclined to start by trying to show that if [itex]f(x)[/itex] is increasing, then:
[tex]1 \times f(n) \leq \integral_{n}{n+1} f(n) \leq 1 \times f(n+1)[/tex]

The sums do correspond to left and right Riemann sums, but that's not going to be part of the proof unless you have a specific theorem about Riemann sums and increasing functions available.
 

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