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Integrable Functions and the Dominated Convergence Theorem · Tutorial 299 of 798

Passing Limits Through Infinite Series

Learn how a single summable sequence can justify termwise limits and convergence of the corresponding infinite sums.

Advanced 11 min read

What You'll Learn

  • State the series limit theorem using counting measure on the nonnegative integers.
  • Check a proposed dominating sequence term by term and verify its sum is finite.
  • Prove convergence of sums using a finite initial segment and a uniformly controlled tail.
  • Apply the theorem to geometric, factorial, and signed series.
  • Identify why pointwise convergence alone does not permit passing a limit through an infinite sum.

Why Infinite Series Need Uniform Tail Control

For a finite sum, passing a limit through the summation sign requires only the convergence of each term. For an infinite series, the terms beyond any fixed index can collectively contribute a substantial amount, even when every fixed term tends to zero. A summable bound prevents this: it controls the tail of every series in the sequence at once, at the same threshold.

We work on \((\mathbb{N}_0,\mathcal{P}(\mathbb{N}_0),\#)\), where \(\mathbb{N}_0=\{0,1,2,\ldots\}\), \(\mathcal{P}(\mathbb{N}_0)\) is the power-set sigma-algebra, and \(\#\) is counting measure. This measure space is sigma-finite, complete, and not finite: each singleton has measure \(1\), and \(\mathbb{N}_0\) is a countable union of finite-measure sets. Every real-valued function on it is measurable. For a function \(u:\mathbb{N}_0\to\mathbb{R}\), integrability means \(\sum_{k=0}^{\infty}|u(k)|<\infty\), and then \(\int u\,d\#=\sum_{k=0}^{\infty}u(k)\). In this space, a statement holds \(\#\)-almost everywhere only if it holds everywhere, since the only set of counting measure zero is the empty set.

The Series Version of Dominated Convergence

Theorem (Dominated convergence for series): For each \(n\ge1\), let \(a_{n,k}\in\mathbb{R}\) for \(k\in\mathbb{N}_0\). Suppose that \(a_{n,k}\to a_k\) as \(n\to\infty\) for every fixed \(k\), and that there is a sequence \(b_k\ge0\) such that \(\sum_{k=0}^{\infty}b_k<\infty\) and \(|a_{n,k}|\le b_k\) for every \(n\ge1\) and every \(k\in\mathbb{N}_0\). Then each series \(\sum_k a_{n,k}\) converges absolutely, \(\sum_k a_k\) converges absolutely, and $$ \lim_{n\to\infty}\sum_{k=0}^{\infty}a_{n,k} =\sum_{k=0}^{\infty}a_k. $$

The theorem is precisely the Dominated Convergence Theorem on the stated counting-measure space: define \(f_n(k)=a_{n,k}\) and \(f(k)=a_k\). The assumptions say \(f_n\to f\) pointwise and \(|f_n|\le g\), where \(g(k)=b_k\) is integrable because its counting-measure integral is \(\sum_k b_k<\infty\). The Dominated Convergence Theorem therefore gives \(\int f_n\,d\#\to\int f\,d\#\), which is the asserted identity of series. The following direct proof also makes explicit how the common summable bound controls the tails.

Proof. Fix \(k\). Since \(a_{n,k}\to a_k\) and \(|a_{n,k}|\le b_k\), continuity of absolute value gives \(|a_k|\le b_k\). Thus \(\sum_k|a_k|\le\sum_kb_k<\infty\). For every \(n\), \(\sum_k|a_{n,k}|\le\sum_kb_k<\infty\) as well, so all the series in the conclusion are absolutely convergent.

Let \(\varepsilon>0\). Since \(\sum_kb_k\) converges, choose \(K\in\mathbb{N}_0\) such that \(\sum_{k>K}b_k<\varepsilon/4\). For the tail, the triangle inequality and the bounds on both \(a_{n,k}\) and \(a_k\) give, for every \(n\),

$$ \sum_{k>K}|a_{n,k}-a_k| \le \sum_{k>K}(|a_{n,k}|+|a_k|) \le 2\sum_{k>K}b_k <\frac{\varepsilon}{2}. $$

There are only finitely many indices \(0\le k\le K\). For each such index, \(a_{n,k}\to a_k\); hence there is an \(N\) such that for \(n\ge N\), \(|a_{n,k}-a_k|<\varepsilon/(2(K+1))\) for every \(0\le k\le K\). Consequently, for \(n\ge N\),

$$ \left|\sum_{k=0}^{\infty}a_{n,k}-\sum_{k=0}^{\infty}a_k\right| \le \sum_{k=0}^{K}|a_{n,k}-a_k|+\sum_{k>K}|a_{n,k}-a_k| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$

This is the definition of convergence of the sums, and proves the claim. \(\square\)

Worked Applications

Worked Example: A Sequence of Geometric Series

For \(n\ge1\) and \(k\in\mathbb{N}_0\), set \(a_{n,k}=\left(\dfrac{n}{2(n+1)}\right)^k\). Find the limit of \(\sum_{k=0}^{\infty}a_{n,k}\) by checking a summable bound and computing the limiting series.

For every \(n\ge1\), \(0<\dfrac{n}{2(n+1)}<\dfrac12\), since \(n<n+1\). Therefore, for every \(k\ge0\),

$$ |a_{n,k}|=\left(\frac{n}{2(n+1)}\right)^k\le\left(\frac12\right)^k. $$

The proposed bound is summable: \(\sum_{k=0}^{\infty}(1/2)^k=1/(1-1/2)=2\). For each fixed \(k\), \(\dfrac{n}{2(n+1)}\to\dfrac12\), so \(a_{n,k}\to 2^{-k}\). The limiting series is \(\sum_{k=0}^{\infty}2^{-k}=2\), and the theorem gives convergence of the original series to \(2\).

We can independently check the result by summing each geometric series. Its ratio \(r_n=n/[2(n+1)]\) is strictly less than \(1\), so \(\sum_{k=0}^{\infty}a_{n,k}=1/(1-r_n)=2(n+1)/(n+2)\). Thus its limit is \(2\). At \(n=1\), the ratio is \(1/4\), and the sum is \(1/(1-1/4)=4/3\), also given by \(2(1+1)/(1+2)=4/3\).

Worked Example: A Factorial Bound and the Exponential Series

For \(n\ge1\) and \(k\in\mathbb{N}_0\), let \(a_{n,k}=\dfrac{(n/(n+1))^k}{k!}\). Show that the sums converge to \(e\).

For \(n\ge1\), \(0<n/(n+1)<1\). Since \(k!\ge1\), the following is a bound for every term, including \(k=0\):

$$ |a_{n,k}|=\frac{(n/(n+1))^k}{k!}\le\frac{1}{k!}. $$

The bound is summable, with \(\sum_{k=0}^{\infty}1/k!=e\), by the exponential series. At each fixed \(k\), \(n/(n+1)\to1\), so \(a_{n,k}\to1/k!\). The limiting series is therefore \(\sum_{k=0}^{\infty}1/k!=e\). The theorem yields \(\lim_{n\to\infty}\sum_{k=0}^{\infty}a_{n,k}=e\).

For a check by direct summation, the exponential series gives \(\sum_{k=0}^{\infty}a_{n,k}=\exp(n/(n+1))\), whose limit is \(\exp(1)=e\). For \(n=1\), the first terms are \(1\), \(1/2\), and \(1/8\): these follow by substituting \(n/(n+1)=1/2\) into the \(k=0,1,2\) terms. Their sum is \(13/8\), and all further terms are nonnegative, consistent with the exact row sum \(\exp(1/2)\).

Worked Example: A Signed Series with a Nonzero Limit

For \(n\ge1\) and \(k\ge1\), define \(a_{n,k}=(-1)^k\dfrac{k}{n+k}2^{-k}\). Show that the row sums converge and compute their limit.

Since \(n\ge1\) and \(k\ge1\), \(0<k/(n+k)<1\). Thus the bound holds term by term for every \(n,k\):

$$ |a_{n,k}|=\frac{k}{n+k}2^{-k}\le2^{-k}. $$

The bound is summable over \(k\ge1\), since \(\sum_{k=1}^{\infty}2^{-k}=1\). For each fixed \(k\), \(k/(n+k)\to0\), so \(a_{n,k}\to0\). The theorem shows that the series in each row is absolutely convergent and that the row sums converge to \(\sum_{k=1}^{\infty}0=0\).

For an explicit tail check, after index \(K=3\) the common bound has sum \(\sum_{k=4}^{\infty}2^{-k}=2^{-3}=1/8\). Consequently, for every \(n\), the absolute value of the tail of the row is at most \(1/8\); this bound is the same for all rows. This control, together with convergence of each of the first four terms, is what justifies the limit of the full sum.

Why the Common Bound Matters

Pointwise convergence of terms alone does not suffice. On the same counting-measure space, for \(n\ge1\) and \(k\ge1\), take \(a_{n,k}=1\) when \(k=n\), and \(a_{n,k}=0\) otherwise. For each fixed \(k\), \(a_{n,k}\to0\): once \(n>k\), the term is zero. However, every row has exactly one nonzero term, so \(\sum_{k=1}^{\infty}a_{n,k}=1\) for every \(n\), whereas the series of pointwise limits is \(0\).

There cannot be a summable common bound for this example. If \(b_k\ge0\) satisfied \(|a_{n,k}|\le b_k\) for every \(n,k\), then choosing \(n=k\) gives \(b_k\ge1\) for every \(k\ge1\), and hence \(\sum_{k=1}^{\infty}b_k=\infty\). The missing hypothesis is exactly the summable bound, which rules out a unit of mass moving to ever-later indices.

Common mistake. It is incorrect to argue, “Each term tends to zero, so the sum tends to zero.” In the spike example, every fixed term does tend to zero, but the sum remains \(1\). The correction is to check a single sequence \(b_k\) that bounds \(|a_{n,k}|\) for every row \(n\), and then verify \(\sum_k b_k<\infty\). Bounds that depend on \(n\), or bounds whose series diverges, do not establish this theorem’s hypothesis.

This series formulation is useful whenever an infinite sum is an integral with respect to counting measure: it turns a question about interchanging a limit and an infinite sum into a question about pointwise convergence and one explicit summable majorant. The distinction between the finite initial segment and the uniformly bounded tail is the essential mechanism.

Check Your Understanding

For each problem, identify the pointwise limit and verify whether the proposed bound is summable before interchanging a limit and a sum.

  1. For \(a_{n,k}=(n/(n+1))^k3^{-k}\), \(k\ge0\), find a summable bound and compute the limit of the row sums.
  2. For \(a_{n,k}=(-1)^k/(n+k)\), \(k\ge1\), does the bound \(b_k=1/k\) suffice for the theorem? Explain which condition it does or does not meet.
  3. For \(a_{n,k}=n^{-1}2^{-k}\), \(k\ge0\), verify a common summable bound and compute the limit of the row sums.
  4. In the spike example, calculate the sum of each row and the series of pointwise limits. Which theorem hypothesis fails?
  5. Explain why a bound \(b_{n,k}\) that is summable in \(k\) for each fixed \(n\) is not, by itself, the common bound required by the series theorem.

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