The integral test applied to the
harmonic series. Since the area under the curve
y = 1/x for
x ∈ [1, ∞) is infinite, the total area of the rectangles must be infinite as well.
Part of a series about

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

Statement of the test 1

Proof 2

Applications 3

Borderline between divergence and convergence 4

References 5
Statement of the test
Consider an integer N and a nonnegative function f defined on the unbounded interval [N, ∞), on which it is monotone decreasing. Then the infinite series

\sum_{n=N}^\infty f(n)
converges to a real number if and only if the improper integral

\int_N^\infty f(x)\,dx
is finite. In other words, if the integral diverges, then the series diverges as well.
If the improper integral is finite, then the proof also gives the lower and upper bounds

\int_N^\infty f(x)\,dx\le\sum_{n=N}^\infty f(n)\le f(N)+\int_N^\infty f(x)\,dx


(1)

for the infinite series.
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 all }x\in[n,\infty)
and

f(n)\le f(x)\quad\text{for all }x\in[N,n].
Hence, for every integer n ≥ N,

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


(2)

and, for every integer n ≥ N + 1,

f(n)=\int_{n1}^{n} f(n)\,dx \le\int_{n1}^n f(x)\,dx.


(3)

By summation over all n from N to some larger integer M, we get from (2)

\int_N^{M+1}f(x)\,dx=\sum_{n=N}^M\underbrace{\int_n^{n+1}f(x)\,dx}_{\le\,f(n)}\le\sum_{n=N}^Mf(n)
and from (3)

\sum_{n=N}^Mf(n)\le f(N)+\sum_{n=N+1}^M\underbrace{\int_{n1}^n f(x)\,dx}_{\ge\,f(n)}=f(N)+\int_N^M f(x)\,dx.
Combining these two estimates yields

\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 bounds in (1) and the result follow.
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

\zeta(1+\varepsilon)=\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}
(cf. Riemann zeta function) converges for every ε > 0, because by the power rule

\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<\infty \quad\text{for all }M\ge1.
From (1) we get the upper estimate

\zeta(1+\varepsilon)=\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}\le\frac{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, whether there are monotone sequences such that f(n) decreases to 0 faster than 1/n but slower than 1/n^{1+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 of infinite series.
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_{k1}(n)\ln_k(n)}


(4)

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_{k1}(n)(\ln_k(n))^{1+\varepsilon}}


(5)

converges for every ε > 0. Here ln_{k} denotes the kfold composition of the natural logarithm defined recursively by

\ln_k(x)= \begin{cases} \ln(x)&\text{for }k=1,\\ \ln(\ln_{k1}(x))&\text{for }k\ge2. \end{cases}
Furthermore, N_{k} denotes the smallest natural number such that the kfold composition is welldefined and ln_{k}(N_{k}) ≥ 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 uparrow notation.
To see the divergence of the series (4) 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 series (5), 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_{k1}(x)(\ln_k(x))^{1+\varepsilon}},
hence

\int_{N_k}^\infty\frac{dx}{x\ln(x)\cdots\ln_{k1}(x)(\ln_k(x))^{1+\varepsilon}} =\frac1{\varepsilon(\ln_k(x))^\varepsilon}\biggr_{N_k}^\infty<\infty
and (1) gives bounds for the infinite series in (5).
References

Knopp, Konrad, "Infinite Sequences and Series", Dover publications, Inc., New York, 1956. (§ 3.3) ISBN 0486601536

Whittaker, E. T., and Watson, G. N., A Course in Modern Analysis, fourth edition, Cambridge University Press, 1963. (§ 4.43) ISBN 0521588073

Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987, ISBN 9723101793
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