Geometric and Telescoping Series as Integrals

From Pointwise Series to Integrals

A pointwise formula for an infinite sum does not by itself determine whether its integral is finite. For instance, a geometric series may converge at almost every point while its sum has infinite integral. The useful feature of nonnegative terms is that the partial sums increase, so the Monotone Convergence Theorem gives an exact way to pass from their integrals to the integral of their pointwise sum.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, completeness, or $\sigma$-finiteness assumption is needed. All functions in the main statements are measurable and take values in $[0,\infty]$, unless a statement explicitly requires finite-valued functions. Integrals and sums of nonnegative quantities are allowed to equal $+\infty$.

Theorem (nonnegative series as integrals): If $g_n:X\to[0,\infty]$ are measurable for $n\ge1$, then $S=\sum_{n=1}^{\infty}g_n$ is measurable and $$ \int_X S\,d\mu=\sum_{n=1}^{\infty}\int_X g_n\,d\mu, $$ with equality in $[0,\infty]$.

Proof. Define the partial sums $S_N=\sum_{n=1}^N g_n$. Each $S_N$ is measurable, $0\le S_N\le S_{N+1}$, and $S_N(x)$ increases to $S(x)$ for every $x$. Thus $S$ is measurable as a pointwise limit of measurable functions. By finite additivity of the nonnegative integral, $\int S_N\,d\mu=\sum_{n=1}^N\int g_n\,d\mu$. The Monotone Convergence Theorem applies to $(S_N)$, giving $\int S\,d\mu=\lim_{N\to\infty}\int S_N\,d\mu=\lim_{N\to\infty}\sum_{n=1}^N\int g_n\,d\mu$. The last limit is the nonnegative series on the right, including when it diverges to $+\infty$. $\square$

Geometric Series, Including the Infinite Case

Proposition (geometric series under the integral): Let $q:X\to[0,1]$ be measurable and define $H:X\to[0,\infty]$ by $H(x)=1/(1-q(x))$ when $q(x)<1$ and $H(x)=+\infty$ when $q(x)=1$. With the convention $q^0=1$, the pointwise identity $$ \sum_{n=0}^{\infty}q^n=H $$ holds everywhere, and $\int_X H\,d\mu=\sum_{n=0}^{\infty}\int_Xq^n\,d\mu$.

Proof. Fix $x\in X$. If $q(x)<1$, the finite geometric-sum identity gives $\sum_{n=0}^Nq(x)^n=(1-q(x)^{N+1})/(1-q(x))$, which tends to $1/(1-q(x))$. If $q(x)=1$, every term is $1$, so the partial sums tend to $+\infty=H(x)$. Each $q^n$ is measurable, and the nonnegative-series theorem just proved gives the integral identity. $\square$

Worked Example: A Constant Geometric Series on a Finite Interval

On $[0,2]$ with Lebesgue measure $m$, calculate the integral of $\sum_{n=0}^{\infty}(1/3)^n$. Every term is a nonnegative measurable constant, and the partial sums increase. The pointwise geometric sum is $1/(1-1/3)=3/2$ at every $x$. Therefore

\int_{[0,2]}\sum_{n=0}^{\infty}(1/3)^n\,dm =\int_{[0,2]}\frac32\,dm =\frac32\,m([0,2]) =\frac32\cdot2=3.

The series-integral identity also computes the same value term by term: $m([0,2])=2$, so $\int_{[0,2]}(1/3)^n\,dm=2(1/3)^n$, and $$ \sum_{n=0}^{\infty}2(1/3)^n =2\cdot\frac{1}{1-1/3} =2\cdot\frac32=3. $$ Check: the pointwise sum is $3/2$ on a set of measure $2$, so its integral is $3/2\cdot2=3$, agreeing with the termwise calculation.

Worked Example: A Geometric Sum with Infinite Integral

On $[0,1]$ with Lebesgue measure $m$, take $q(x)=x$. For $0\le x<1$, the geometric sum is $1/(1-x)$, while at $x=1$ it is $+\infty$. The terms $x^n$ are nonnegative and measurable, so the nonnegative-series theorem applies, including at the endpoint. Directly integrating the terms gives

\int_{[0,1]}\sum_{n=0}^{\infty}x^n\,dm =\sum_{n=0}^{\infty}\int_0^1x^n\,dx =\sum_{n=0}^{\infty}\frac{1}{n+1} =+\infty.

To verify the divergence, group the harmonic series as follows. For each integer $k\ge0$, the terms with $2^k\le j<2^{k+1}$ contribute at least $2^k\cdot 1/2^{k+1}=1/2$ to $\sum_{j=1}^{\infty}1/j$. There are infinitely many such disjoint groups, so their partial totals are unbounded. As an independent check using the pointwise sum, for $0<\varepsilon<1$, $$ \int_0^{1-\varepsilon}\frac{1}{1-x}\,dx =-\ln(\varepsilon). $$ As $\varepsilon$ decreases to $0$, $-\ln(\varepsilon)$ increases without bound. Thus the integral of the geometric sum on $[0,1]$ is infinite, as the series of term integrals showed.

Check: the sum is finite at every $x<1$, but that does not imply a finite integral. The endpoint value $+\infty$ alone is not the reason for divergence: a single point has Lebesgue measure zero. The unbounded integrals over $[0,1-\varepsilon]$ verify the divergence directly.

Telescoping Nonnegative Terms

A telescoping series can be built from the successive differences of an increasing sequence. To avoid undefined subtraction, the sequence in the following result is required to be finite-valued at every point. Its pointwise limit is allowed to be infinite.

Proposition (finite-valued telescoping functions): Let $v_0=0$, and for every $n\ge1$ let $v_n:X\to[0,\infty)$ be finite-valued and measurable, with $v_{n-1}(x)\le v_n(x)$ for every $x\in X$. Define $g_n=v_n-v_{n-1}$. Then each $g_n$ is nonnegative and measurable, $\sum_{n=1}^N g_n=v_N$ everywhere, and, with $v(x)=\lim_{N\to\infty}v_N(x)\in[0,\infty]$, $$ \int_X v\,d\mu=\lim_{N\to\infty}\int_Xv_N\,d\mu =\sum_{n=1}^{\infty}\int_Xg_n\,d\mu. $$

Proof. Since both $v_n(x)$ and $v_{n-1}(x)$ are finite, $g_n(x)$ is defined as an ordinary real difference. The assumed pointwise inequality gives $g_n(x)\ge0$; measurability follows from measurability of the finite-valued functions and their difference. For each $N$, cancellation of the finite sum gives $\sum_{n=1}^Ng_n=(v_1-v_0)+(v_2-v_1)+\cdots+(v_N-v_{N-1})=v_N$ at every point. The sequence $v_N$ increases pointwise to $v$, so the Monotone Convergence Theorem gives $\int v\,d\mu=\lim_N\int v_N\,d\mu$. Also, finite additivity gives $\int v_N\,d\mu=\sum_{n=1}^N\int g_n\,d\mu$. Taking the limit of these partial sums proves the last equality, with $+\infty$ permitted. $\square$

Worked Example: A Finite Telescoping Total on the Unit Interval

On $[0,1]$ with Lebesgue measure, define $v_0(x)=0$ and $v_N(x)=1-(1-x)^N$ for $N\ge1$. Each $v_N$ is measurable and finite-valued. Since $0\le1-x\le1$, the powers $(1-x)^N$ decrease as $N$ increases, so $v_N$ increases. Its successive differences are $$ g_1(x)=x,\qquad g_n(x)=x(1-x)^{n-1}\quad(n\ge2). $$ These are nonnegative, and the finite telescoping identity gives $\sum_{n=1}^Ng_n(x)=1-(1-x)^N$.

For every $x>0$, $(1-x)^N\to0$, including $x=1$; for $x=0$, $v_N(0)=0$. Hence the pointwise limit $v$ equals $1$ on $(0,1]$ and $0$ at $0$. Its integral is $1$, since the exceptional singleton $\{0\}$ has Lebesgue measure zero. We can check by integrating each term: for $n\ge1$, the formula $g_n(x)=x(1-x)^{n-1}$ also holds when $n=1$, and the substitution $t=1-x$ yields $$ \int_0^1 x(1-x)^{n-1}\,dx =\int_0^1(1-t)t^{n-1}\,dt =\frac1n-\frac1{n+1} =\frac{1}{n(n+1)}. $$ Therefore $$ \sum_{n=1}^{\infty}\int_0^1g_n\,dm =\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right)=1. $$ Check: the first three term integrals are $1/2$, $1/6$, and $1/12$, whose sum is $3/4$. Independently, $\int_0^1v_3(x)\,dx=1-\int_0^1(1-x)^3\,dx=1-1/4=3/4$, matching the three-term sum.

Worked Example: A Telescoping Sequence with Infinite Total

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}$ be all subsets of $\mathbb{N}$, and let $\mu$ be counting measure. Set $v_0(k)=0$ and $v_N(k)=1$ if $k\le N$, while $v_N(k)=0$ if $k>N$. For each fixed $k$, this is a nondecreasing sequence of finite-valued measurable functions. Its differences are $g_n(k)=1$ if $k=n$ and $g_n(k)=0$ otherwise. Thus $\sum_{n=1}^\infty g_n(k)=1$ for every $k$, and the pointwise limit $v$ is the constant function $1$.

Counting measure integrates a nonnegative function by summing its values over $\mathbb{N}$. Consequently $\int v\,d\mu=\sum_{k=1}^{\infty}1=+\infty$. The partial integrals are $\int v_N\,d\mu=N$, because $v_N$ equals $1$ at exactly the $N$ points $1,\ldots,N$ and $0$ elsewhere. On the series side, $\int g_n\,d\mu=1$ for each $n$, so $\sum_{n=1}^{\infty}\int g_n\,d\mu=\sum_{n=1}^{\infty}1=+\infty$. Check: the partial integral of $v_N$ and the partial sum of the first $N$ term integrals are both $N$; both tend to $+\infty$.

Common Pitfalls and Why the Method Matters

A common error is to infer that an integral is finite merely because a pointwise sum is finite almost everywhere. In the geometric example $x^n$ on $[0,1]$, the sum is finite for every $x<1$, yet its integral is infinite. The correction is to compute the integral of the sum or the sum of the term integrals, using the nonnegative-series theorem; pointwise finiteness is not an integrability test.

A second error is to telescope extended-valued functions by subtracting infinities. If $v_1(x)=+\infty$ and $v_2(x)=+\infty$, then $v_2(x)-v_1(x)$ is undefined, not zero. The telescoping proposition therefore assumes every $v_n$ is finite-valued. The limit $v=\lim_n v_n$ may still be $+\infty$; only the individual functions whose differences are taken must be finite-valued.

For both geometric and telescoping series, nonnegativity ensures increasing partial sums. Monotone convergence then identifies the integral of the pointwise limit with the limit of the partial-sum integrals, even when that value is infinite. Without that structure, these arguments do not justify exchanging a limit and an integral.

Check Your Understanding

For each question, state whether the resulting integral or series is finite, and identify the nonnegative partial sums to which monotone convergence applies.

On $[0,1]$ with Lebesgue measure, compute $\int_0^1\sum_{n=0}^{\infty}(x/2)^n\,dm$ using both the pointwise geometric sum and the sum of term integrals.
On $[0,1]$, let $g_n(x)=2^{-n}$ for $n\ge1$. Find the pointwise sum and its integral.
For finite-valued measurable $v_n$ with $0=v_0\le v_1\le v_2\le\cdots$, explain why $g_n=v_n-v_{n-1}$ is nonnegative and why its partial sums equal $v_N$.
On counting measure on $\mathbb{N}$, let $v_N(k)=1$ for $k\le N$ and $v_N(k)=0$ for $k>N$. Compute $\int v_N\,d\mu$ and the integral of its pointwise limit.
Why is the expression $v_n-v_{n-1}$ not valid as a general definition when both functions may take the value $+\infty$?