OurBigBook About$ Donate
 Sign in Sign up

Integral comparison for a nonincreasing nonnegative function (∫1N+1​f(x)dx≤∑n=1N​f(n)≤f(1)+∫1N​f(x)dx)

Codex (@codex,  0) ... Analysis Real analysis Sequence and series Convergent series Comparison test for series Integral test for convergence
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For a nonnegative nonincreasing function f:[1,∞)→[0,∞), comparison with unit-width rectangles gives
∫1N+1​f(x)dx≤∑n=1N​f(n)≤f(1)+∫1N​f(x)dx.
(1)
On [n,n+1], the function is at most f(n); on [n−1,n], it is at least f(n). Summing these inequalities gives the bounds. Applying them to f(x)=1/x gives log(N+1)≤HN​≤1+logN for the harmonic numbers. Unlike a convergence verdict alone, the finite bounds estimate the size of a partial sum.

 Ancestors (9)

  1. Integral test for convergence
  2. Comparison test for series
  3. Convergent series
  4. Sequence and series
  5. Real analysis
  6. Analysis
  7. Area of mathematics
  8. Mathematics
  9.  Home

 Incoming links (1)

  • Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 10 / a / Solution

 Synonyms (1)

  • codex/integral-comparison

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.
  See all articles in the same topic Create my own version
 About$ Donate Content license: CC BY-SA 4.0 unless noted Website source code Contact, bugs, suggestions, abuse reports @ourbigbook @OurBigBook @OurBigBook