Integral test for convergence explained

In mathematics, the integral test for convergence is a method used to test infinite series of monotonous terms for convergence. It was developed by Colin Maclaurin and Augustin-Louis Cauchy and is sometimes known as the Maclaurin–Cauchy test.

Statement of the test

Consider an integer and a function defined on the unbounded interval, on which it is monotone decreasing. Then the infinite series

\sum ^{inf ty} _{n = N} f(n)

converges to a real number if and only if the improper integral

\int ^{inf ty} _{N} f(x)dx

is finite. In particular, if the integral diverges, then the series diverges as well.

Remark

If the improper integral is finite, then the proof also gives the lower and upper bounds

for the infinite series.

Note that if the function

f(x) is increasing, then the function - f(x) is decreasing and the above theorem applies.

Many textbooks require the function

f to be positive,^{[1] [2] [3]} but this condition is not really necessary. Since when f is negative and decreasing, both \sum ^{inf ty} _{n = N} f(n) and \int ^{inf ty} _{N} f(x)dx diverges as discussed in Mathematics Stack Exchange.^[4]

Proof

The proof basically uses the comparison test, comparing the term with the integral of over the intervals and, respectively.

The monotonous function

f is continuous almost everywhere. To show this, let D = \{ x\in [N, inf ty)\mid f isdiscontinuousat x\} . For every x\in D , there exists by the density of Q a c(x)\in Q so that c(x)\in\left[\lim _{y\downarrow} f(y), \lim _{y\uparrow} f(y)\right] . Note that this set contains an open non-empty interval precisely if f is discontinuous at x . We can uniquely identify c(x) as the rational number that has the least index in an enumeration N\to Q and satisfies the above property. Since f is monotone, this defines an injective mapping c:D\to Q, x\mapsto c(x) and thus D is countable. It follows that f is continuous almost everywhere. This is sufficient for Riemann integrability.^[5]

Since is a monotone decreasing function, we know that

f(x)\le f(n) forall x\in[n, inf ty)

and

f(n)\le f(x) forall x\in[N,n].

Hence, for every integer,

and, for every integer,

By summation over all from to some larger integer, we get from

\int ^{M +1} _{N} f(x)dx = \sum ^{n +1} _{n} f(x)dx} _{\lef(n)} \le\sum ^{Mf(n)} _{n = N}

and from

\sum ^{Mf(n)\le} _{n = N +1} f(N) + \sum ^{n} _{n -1} f(x)dx} _{\gef(n)} = f(N) + \int ^{M} _{N} f(x)dx.

Combining these two estimates yields

\int ^{M +1} _{N} f(x)dx\le\sum ^{Mf(n)\le} _{n = N} f(N) + \int ^{M} _{N} f(x)dx.

Letting tend to infinity, the bounds in and the result follow.

Applications

The harmonic series

\sum ^{inf ty} _{n =1} (1)/(n) diverges because, using the natural logarithm, its antiderivative, and the fundamental theorem of calculus, we get\int ^{M} _{1} (1)/(ndn) = ln n r | ^{M} _{1} = ln M \to inf ty for M\to inf ty. On the other hand, the series\zeta( 1+ \varepsilon) = \sum ^{inf ty} _{n =1} (1 {n)/(^{1+\varepsilon}) }(cf. Riemann zeta function)converges for every, because by the power rule\int ^{M (1 {n)/(^{1+\varepsilon})} _{1} }\,dn= \left. -\frac 1 \right|_1^M=\frac 1 \varepsilon \left(1-\frac 1 \right)\le \frac 1 \varepsilon < \infty\quad\textM\ge1.From we get the upper estimate\zeta( 1+ \varepsilon) = \sum ^{inf ty} _{n =1} (1)/(n ^{1+\varepsilon}) \le (1+ \varepsilon)/(\varepsilon,) which can be compared with some of the particular values of Riemann zeta function.

Borderline between divergence and convergence

The above examples involving the harmonic series raise the question of whether there are monotone sequences such that decreases to 0 faster than but slower than in the sense that

\lim _{n\toinfty} (f(n))/(1 /n) =0 and \lim _{n\toinfty} (f(n))/(1 /n ^{1+\varepsilon}) =inf ty for every, and whether the corresponding series of the still diverges. Once such a sequence is found, a similar question can be asked with taking the role of, and so on. In this way it is possible to investigate the borderline between divergence and convergence of infinite series.

Using the integral test for convergence, one can show (see below) that, for every natural number, the seriesstill diverges (cf. proof that the sum of the reciprocals of the primes diverges for) butconverges for every . Here denotes the -fold composition of the natural logarithm defined recursively by

ln _{k(x)= \begin{cases} ln(x)&fork=1,\\ ln(ln} _{k-1} (x))& for k\ge 2 . \end{cases} Furthermore, denotes the smallest natural number such that the -fold composition is well-defined and, i.e.N _{k\ge} \underbrace{e ^{e ^{⋅ ^{⋅ ^{e}}}} }_=e \uparrow\uparrow kusing tetration or Knuth's up-arrow notation.

To see the divergence of the series using the integral test, note that by repeated application of the chain rule

(d)/(dx) ln _{k+1} (x) = (d)/(dx) ln ( ln _{k(x)) = (1 { ln)/(^{k(x)} (d)/(dx)})} ln _{k(x) = … = (1 {x ln (x) … ln)/(_{k(x)},})} hence\int ^{inf ty (dx)/(x ln (x) … ln _{k(x)})} _{N _{k}} =ln _{k+1} (x) r | ^{inf ty =inf ty.} _{N _{k}} To see the convergence of the series, note that by the power rule, the chain rule and the above result- (d)/(dx) (1 {\varepsilon( ln)/(_{k(x))}) ^{\varepsilon} = (1 {( ln)/(_{k(x))})} ^{1+\varepsilon} }\frac\ln_k(x)=\cdots=\frac,hence\int ^{inf ty (dx)/(x ln (x) … ln _{k -1} (x)( ln ^{1+ \varepsilon} _{k(x))})} _{N _{k}} =- (1 {\varepsilon( ln)/(_{k(x))}) ^{inf ty< inf ty} ^{N _{k}} and gives bounds for the infinite series in .

See also

• Convergence tests
• Convergence (mathematics)
• Direct comparison test
• Dominated convergence theorem
• Euler-Maclaurin formula
• Limit comparison test
• Monotone convergence theorem

References

• Knopp, Konrad, "Infinite Sequences and Series", Dover Publications, Inc., New York, 1956. (§ 3.3)
• Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43)
• Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987,

1. Book: Stewart . James . Clegg . Daniel . Watson . Saleem . Calculus: Metric Version . 2021 . Cengage . 9780357113462 . 9.
2. Book: Wade . William . An Introduction to Analysis . 2004 . Pearson Education . 9780131246836 . 3.
3. Book: Thomas . George . Hass . Joel . Heil . Christopher . Weir . Maurice . Zuleta . José Luis . Thomas' Calculus: Early Transcendentals . 2018 . Pearson Education . 9781292253114 . 14.
4. Web site: Why does it have to be positive and decreasing to apply the integral test? . savemycalculus . Mathematics Stack Exchange . 2020-03-11.
5. 0002-9890. 43. 7. 396–398. Brown. A. B.. A Proof of the Lebesgue Condition for Riemann Integrability. The American Mathematical Monthly. September 1936. 2301737. 10.2307/2301737.