integral test for convergence

Consider an integer {{math|N}} and a function {{math|f}} defined on the unbounded interval {{closed-open|N, ∞}}, on which it is monotone decreasing. Then the infinite series

:$\backslash sum\_\{n=N\}^\backslash infty\; f(n)$

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

:$\backslash int\_N^\backslash infty\; f(x)\backslash ,dx$

is finite. In particular, 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

{{NumBlk|:|$\backslash int\_N^\backslash infty\; f(x)\backslash ,dx\backslash le\backslash sum\_\{n=N\}^\backslash infty\; f(n)\backslash le\; f(N)+\backslash int\_N^\backslash infty\; f(x)\backslash ,dx$|{{EquationRef|1}}}}

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,{{cite book |last1=Stewart |first1=James |last2=Clegg |first2=Daniel |last3=Watson |first3=Saleem |title=Calculus: Metric Version |date=2021 |publisher=Cengage |isbn=9780357113462 |edition=9}}{{cite book |last1=Wade |first1=William |title=An Introduction to Analysis |date=2004 |publisher=Pearson Education |isbn=9780131246836 |edition=3}}{{cite book |last1=Thomas |first1=George |last2=Hass |first2=Joel |last3=Heil |first3=Christopher |last4=Weir |first4=Maurice |last5=Zuleta |first5=José Luis |title=Thomas' Calculus: Early Transcendentals |date=2018 |publisher=Pearson Education |isbn=9781292253114 |edition=14}} but this condition is not really necessary, since when $f$ is negative and decreasing both $\backslash sum\_\{n=N\}^\backslash infty\; f(n)$ and $\backslash int\_N^\backslash infty\; f(x)\backslash ,dx$ diverge.{{cite web |url=https://math.stackexchange.com/q/3577379 |title=Why does it have to be positive and decreasing to apply the integral test? |author=savemycalculus |website=Mathematics Stack Exchange |access-date=2020-03-11}}{{better source needed|date=August 2024}}

The proof 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.

The monotonic function $f$ is continuous almost everywhere. To show this, let

:$D=\backslash \{\; x\backslash in\; [N,\backslash infty)\backslash mid\; f\backslash text\{\; is\; discontinuous\; at\; \}\; x\backslash \}$

For every $x\backslash in\; D$, there exists by the density of $\backslash mathbb\; Q$, a $c(x)\backslash in\backslash mathbb\; Q$ so that $c(x)\backslash in\backslash left[\backslash lim\_\{y\backslash downarrow\; x\}\; f(y),\; \backslash lim\_\{y\backslash uparrow\; x\}\; f(y)\backslash right]$.

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 $\backslash mathbb\; N\backslash to\backslash mathbb\; Q$ and satisfies the above property. Since $f$ is monotone, this defines an injective mapping $c:D\backslash to\backslash mathbb\; Q,\; x\backslash mapsto\; c(x)$ and thus $D$ is countable. It follows that $f$ is continuous almost everywhere. This is sufficient for Riemann integrability.{{Cite journal| issn = 0002-9890| volume = 43

| issue = 7| pages = 396–398| last = Brown| first = A. B.| title = A Proof of the Lebesgue Condition for Riemann Integrability| journal = The American Mathematical Monthly| date = September 1936| jstor = 2301737| doi = 10.2307/2301737}}

Since {{math|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 {{math|n ≥ N}},

{{NumBlk|:|$$

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

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

=f(n)|{{EquationRef|2}}}}

and, for every integer {{math|n ≥ N + 1}},

{{NumBlk|:|$$

f(n)=\int_{n-1}^{n} f(n)\,dx

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

|{{EquationRef|3}}}}

By summation over all {{math|n}} from {{math|N}} to some larger integer {{math|M}}, we get from ({{EquationNote|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 ({{EquationNote|3}})

:$$

\begin{align}

\sum_{n=N}^Mf(n)&=f(N)+\sum_{n=N+1}^Mf(n)\\

&\leq f(N)+\sum_{n=N+1}^M\underbrace{\int_{n-1}^n f(x)\,dx}_{\ge\,f(n)}\\

&=f(N)+\int_N^M f(x)\,dx.

\end{align}

Combining these two estimates yields

:$\backslash int\_N^\{M+1\}f(x)\backslash ,dx\backslash le\backslash sum\_\{n=N\}^Mf(n)\backslash le\; f(N)+\backslash int\_N^M\; f(x)\backslash ,dx.$

Letting {{math|M}} tend to infinity, the bounds in ({{EquationNote|1}}) and the result follow.

The harmonic series

:$$

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

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

:$$

\int_1^M \frac 1 n\,dn = \ln n\Bigr|_1^M = \ln M \to\infty

\quad\text{for }M\to\infty.

On the other hand, the series

:$$

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

(cf. Riemann zeta function)

converges for every {{math|ε > 0}}, because by the power rule

:$$

\int_1^M\frac1{n^{1+\varepsilon}}\,dn

= \left. -\frac 1{\varepsilon n^\varepsilon} \right|_1^M=

\frac 1 \varepsilon \left(1-\frac 1 {M^\varepsilon}\right)

\le \frac 1 \varepsilon < \infty

\quad\text{for all }M\ge1.

From ({{EquationNote|1}}) we get the upper estimate

:$$

\zeta(1+\varepsilon)=\sum_{n=1}^\infty \frac 1 {n^{1+\varepsilon}} \le \frac{1 + \varepsilon}\varepsilon,

which can be compared with some of the particular values of Riemann zeta function.

The above examples involving the harmonic series raise the question of whether there are monotone sequences such that {{math|f(n)}} decreases to 0 faster than {{math|1/n}} but slower than {{math|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 {{math|ε > 0}}, and whether the corresponding series of the {{math|f(n)}} still diverges. Once such a sequence is found, a similar question can be asked with {{math|f(n)}} taking the role of {{math|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 {{math|k}}, the series

{{NumBlk|:|$$

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

|{{EquationRef|4}}}}

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

{{NumBlk|:|$$

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

|{{EquationRef|5}}}}

converges for every {{math|ε > 0}}. Here {{math|ln_k}} denotes the {{math|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, {{math|N_k}} denotes the smallest natural number such that the {{math|k}}-fold composition is well-defined and {{math|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 up-arrow notation.

To see the divergence of the series ({{EquationNote|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 ({{EquationNote|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_{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

and ({{EquationNote|1}}) gives bounds for the infinite series in ({{EquationNote|5}}).

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

• 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}}
• Ferreira, Jaime Campos, Ed Calouste Gulbenkian, 1987, {{ISBN|972-31-0179-3}}

{{Calculus topics}}

Category:Augustin-Louis Cauchy

Category:Integral calculus

Category:Convergence tests

Category:Articles containing proofs