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ZixuanZhang
ZixuanZhang
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Integral test

For continuous decreasing and : converges exactly when exists, and for some .

Proposition 1.34 (Integral Test)

Suppose is a continuous decreasing function (so it is integrable in for each [we will see this later]). Let for each .

Then

Furthermore, as ,

Remark. The RHS is an improper integral, which will be discussed later. The last part tells us that the integral is a good approximation for the series (if it converges), or the rate of divergence (if it diverges).

Proof

Comparing each term with the area under the graph over an interval of length one gives

If converges with sum , then , so is increasing and bounded above, hence converges.

Conversely, if the integrals converge then they are bounded, so is increasing and bounded: a monotone bounded sequence converges.

For the error term, satisfies and , so decreases to some limit ; since , also .

Examples

  • converges if and only if : with ,

which has a finite limit exactly for . This also explains the divergence of the harmonic series ().

  • diverges, since under the substitution .

  • converges, since under the same substitution.

Related

Stated in