Integral test for convergence

From Wikipedia, the free encyclopedia

In mathematics, the integral test for convergence is a method used to test infinite series of non-negative terms for convergence. An early form of the test of convergence was developed in India by Madhava in the 14th century, and by his followers at the Kerala School. In Europe, it was later developed by Maclaurin and Cauchy and is sometimes known as the Maclaurin–Cauchy test.

Contents

[edit] Statement of the test

Consider an integer N and a non-negative monotone decreasing function f defined on the unbounded interval [N, ∞). Then the series

\sum_{n=N}^\infty f(n)

converges if and only if the integral

\int_N^\infty f(x)\,dx

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

[edit] Proof

The proof basically uses the comparison test, comparing the term f(n) with the integral of f over the intervals [n − 1, n] and [n, n + 1], respectively.

Since f is a monotone decreasing function, we know that

f(x)\le f(n)\quad\text{for }x\in[n,\infty)

and

f(n)\le f(x)\quad\text{for }x\in[N,n],

hence for every n larger than N

\int_n^{n+1} f(x)\,dx \le\int_{n}^{n+1} f(n)\,dx =f(n) =\int_{n-1}^{n} f(n)\,dx \le\int_{n-1}^n f(x)\,dx.

Since the lower estimate is also valid for f(N), we get by summation over all n from N to some larger integer M

\int_N^{M+1}f(x)\,dx\le\sum_{n=N}^Mf(n)\le f(N)+\int_N^M f(x)\,dx.

Letting M tend to infinity, the result follows.

[edit] Applications

The harmonic series

\sum_{n=1}^\infty \frac1n

diverges because, using the natural logarithm, its derivative, and the fundamental theorem of calculus, we get

\int_1^M\frac1x\,dx=\ln x\Bigr|_1^M=\ln M\to\infty \quad\text{for }M\to\infty.

Contrary, the series

\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}

(cf. Riemann zeta function) converges for every ε > 0, because

\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon \quad\text{for all }M\ge1.

[edit] Borderline between divergence and convergence

The above examples involving the harmonic series raise the question, whether there are monotone sequences such that f(n) decreases to 0 faster than 1/n but slower than 1/n1+ε in the sense that

\lim_{n\to\infty}\frac{f(n)}{1/n}=0 \quad\text{and}\quad \lim_{n\to\infty}\frac{f(n)}{1/n^{1+\varepsilon}}=\infty

for every ε > 0, and whether the corresponding series of the f(n) still diverges. Once such a sequence is found, a similar question can be asked with f(n) taking the role of 1/n, and so on. In this way it is possible to investigate the borderline between divergence and convergence.

Using the integral test for convergence, one can show (see below) that, for every natural number k, the series

\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots \ln_{k-1}(n)\ln_k(n)}

still diverges (cf. proof that the sum of the reciprocals of the primes diverges for k = 1) but

\sum_{n=N_k}^\infty\frac1{n\ln(n)\ln_2(n)\cdots\ln_{k-1}(n)(\ln_k(n))^{1+\varepsilon}}

converges for every ε > 0. Here lnk denotes the k-fold composition of the natural logarithm defined recursively by

\ln_k(x)= \begin{cases} \ln(x)&\text{for }k=1,\\ \ln(\ln_{k-1}(x))&\text{for }k\ge2. \end{cases}

Furthermore, Nk denotes the smallest natural number such that the k-fold composition is well-defined and lnk Nk ≥ 1, i.e.

N_k\ge \underbrace{e^{e^{\cdot^{\cdot^{e}}}}}_{k\ e'\text{s}}=e \uparrow\uparrow k

using tetration or Knuth's up-arrow notation.

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

\frac{d}{dx}\ln_{k+1}(x) =\frac{d}{dx}\ln(\ln_k(x)) =\frac1{\ln_k(x)}\frac{d}{dx}\ln_k(x) =\cdots =\frac1{x\ln(x)\cdots\ln_k(x)},

hence

\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_k(x)} =\ln_{k+1}(x)\bigr|_{N_k}^\infty=\infty.

To see the convergence of the second series, note that by the power rule, the chain rule and the above result

-\frac{d}{dx}\frac1{\varepsilon(\ln_k(x))^\varepsilon} =\frac1{(\ln_k(x))^{1+\varepsilon}}\frac{d}{dx}\ln_k(x) =\cdots =\frac{1}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}},

hence

\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_{k-1}(x)(\ln_k(x))^{1+\varepsilon}} =-\frac1{\varepsilon(\ln_k(x))^\varepsilon}\biggr|_{N_k}^\infty<\infty.

[edit] References

  • Knopp, Konrad, "Infinite Sequences and Series", Dover publications, Inc., New York, 1956. (§ 3.3) ISBN 0-486-60153-6
  • Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43) ISBN 0-521-58807-3