Partial Sums as a Sequence to Study
In the previous tutorial, an infinite series was defined through the sequence of its partial sums. That definition makes the partial sums more than temporary bookkeeping: they are the sequence whose behavior determines the series. This tutorial develops ways to read and manipulate that sequence. In particular, consecutive partial sums recover individual terms, differences of partial sums recover blocks of terms, and useful finite-sum identities can be written directly in terms of partial sums.
For a sequence \((a_n)_{n=1}^{\infty}\), continue to write \(S_N=\sum_{n=1}^{N}a_n\), and set \(S_0=0\) as a convenient convention. The definition of \(S_N\) immediately gives a difference formula. It also works in reverse: any proposed sequence of partial sums, when started at zero, determines exactly one sequence of terms.
Proof. Starting from \(S_0=0\) and \(a_n=S_n-S_{n-1}\), the \(N\)th partial sum is
Every intermediate term cancels in this finite sum. Conversely, if \(S_N=\sum_{n=1}^{N}a_n\), then for \(n\geq2\),
For \(n=1\), the same formula holds because \(S_1=a_1\) and \(S_0=0\). Thus the terms and partial sums determine each other through consecutive differences. \(\square\)
This correspondence is useful in both directions. If the terms have a simple pattern, summing them may reveal a manageable formula for \(S_N\). If a formula for \(S_N\) is known or guessed, its consecutive differences reveal the corresponding terms. The latter approach is often the cleanest way to verify a proposed finite-sum formula.
Worked Example: Finding Terms from a Proposed Partial-Sum Formula
Suppose a sequence is proposed to have partial sums \(S_N=N^3\), with \(S_0=0\). The corresponding terms must be the consecutive differences:
For example, \(a_1=1\), \(a_2=7\), and \(a_3=19\). Their first three partial sums are \(1\), \(1+7=8\), and \(1+7+19=27\), which agree with \(1^3\), \(2^3\), and \(3^3\). The theorem verifies the formula for every \(N\): summing the consecutive differences gives \(S_N=N^3\). This example illustrates how a formula for partial sums can be checked through the terms it implies.
Differences of Partial Sums and Blocks
The same subtraction that extracts one term can extract several consecutive terms. If \(0\leq M<N\), then \(S_N\) includes the first \(M\) terms as well as the terms from \(M+1\) through \(N\). Subtracting \(S_M\) removes the shared initial terms. This gives an exact identity for every finite block.
Proof. By definition, \(S_N=\sum_{n=1}^{N}a_n\). If \(M\geq1\), also \(S_M=\sum_{n=1}^{M}a_n\), so subtraction cancels the terms with indices \(1\) through \(M\) and leaves exactly the terms with indices \(M+1\) through \(N\). If \(M=0\), then \(S_M=S_0=0\), and the identity reduces to the definition \(S_N=\sum_{n=1}^{N}a_n\). These cases cover every permitted \(M\). \(\square\)
The identity is especially helpful when a sum begins at an index other than \(1\). It also shows why a series’ partial sums encode all of its finite blocks: once \(S_M\) and \(S_N\) are known, the sum from \(M+1\) to \(N\) is determined. The result concerns finite sums and needs no assumption that the infinite series converges.
Worked Example: Summing Consecutive Odd Numbers
Take \(a_n=2n-1\). The first \(N\) terms are the odd numbers \(1,3,\ldots,2N-1\). The formula \(S_N=N^2\) can be verified by checking its consecutive differences:
and \(S_0=0\). The Recovering Terms from Partial Sums Theorem therefore gives \(S_N=N^2\). For example, the sum of the odd terms from \(n=4\) through \(n=7\) is
Directly, these four terms are \(7+9+11+13=40\), confirming the block calculation.
A Finite Summation-by-Parts Identity
Partial sums can also simplify finite sums in which each term is multiplied by a changing weight. The resulting identity is the discrete counterpart of integration by parts, but here it is purely an algebraic statement about finite sums. It is sometimes called summation by parts or Abel’s summation formula.
Proof. Since \(a_k=A_k-A_{k-1}\), the left-hand side can be expanded as
In the second sum, \(A_0=0\), so its nonzero-index terms can be written as
Substituting this expression into the difference gives
as claimed. When \(N=1\), the sum from \(k=1\) to \(N-1\) is empty and is interpreted as zero; the identity then reads \(a_1b_1=A_1b_1\), which holds because \(A_1=a_1\). \(\square\)
The formula replaces the individual terms \(a_k\) by their accumulated sums \(A_k\), while the weights appear through their successive differences. It is useful when partial sums of the \(a_k\) are easier to control than the individual weighted terms. The identity itself is finite; applying it to an infinite series requires additional arguments about limits, which will be developed in the study of convergence.
Worked Example: Alternating Signs with Decreasing Weights
Let \(a_k=(-1)^{k-1}\) and \(b_k=1/k\). The partial sums \(A_k\) are \(1\) for odd \(k\) and \(0\) for even \(k\). Finite summation by parts gives
Only odd indices contribute to the last sum. For \(N=4\), \(A_4=0\), \(A_1=A_3=1\), and \(A_2=0\). Hence
Direct calculation gives \(1-\frac12+\frac13-\frac14=\frac7{12}\), as required. For \(N=5\), the endpoint term is \(A_5/5=1/5\), while the contributing differences are still those at \(k=1\) and \(k=3\). The identity gives \(\frac12+\frac1{12}+\frac15=\frac{47}{60}\), and direct calculation gives \(1-\frac12+\frac13-\frac14+\frac15=\frac{47}{60}\). These checks illustrate how the endpoint term changes with \(N\), even when the earlier differences remain the same.
Nonnegative Terms and the Shape of Partial Sums
When every term is nonnegative, the partial sums have a particularly simple shape: they cannot decrease. This connects the signs of the terms to the order behavior of their partial sums. If those partial sums are also bounded above, completeness of the real numbers guarantees that they approach a finite limit.
Proof. For every \(N\geq1\),
so \(S_{N+1}\geq S_N\), and the sequence is nondecreasing. Now suppose it is bounded above. Let \(L\) be the supremum of the set \(\{S_N:N\geq1\}\). Given \(\varepsilon>0\), the definition of supremum ensures that some \(S_{N_0}\) satisfies \(L-\varepsilon<S_{N_0}\); otherwise \(L-\varepsilon\) would be an upper bound smaller than \(L\). For every \(N\geq N_0\), monotonicity and the fact that \(L\) is an upper bound give
Therefore \(|S_N-L|<\varepsilon\) for every \(N\geq N_0\), which proves \(S_N\to L\). Finally, suppose that \((S_N)\) is not bounded above. Given any \(C\in\mathbb{R}\), choose \(N_0\) with \(S_{N_0}>C\). Since the sequence is nondecreasing, \(S_N\geq S_{N_0}>C\) whenever \(N\geq N_0\). This proves the final assertion. \(\square\)
The nonnegativity hypothesis matters: without it, partial sums may move both up and down, so boundedness alone does not guarantee a limit. Even with nonnegative terms, the theorem gives a useful structural conclusion rather than a shortcut around checking the hypotheses: first establish that the terms are nonnegative, then determine whether the partial sums have an upper bound.
Worked Example: A Bounded Sequence of Partial Sums
Let \(a_n=3^{-n}\), so every term is positive. The finite geometric-sum calculation gives
For a direct check, the formula gives \(S_1=\frac12(1-\frac13)=\frac13\), the first term. At \(N=2\), it gives \(\frac12(1-\frac19)=\frac49=\frac13+\frac19\). For every \(N\), \(0<3^{-N}\), so \(S_N<1/2\). The partial sums are increasing because their successive differences are \(a_{N+1}>0\), and they are bounded above by \(1/2\). The Bounded Increasing Partial Sums Theorem therefore ensures that they converge to a finite real number. In fact, the displayed formula shows that their limit is \(1/2\).
Why Partial-Sum Identities Matter
Partial sums make the finite structure of an infinite series explicit. Their consecutive differences give the terms; differences between two partial sums give finite blocks; and summation by parts reorganizes a weighted finite sum in terms of cumulative values. These are exact identities before any limiting argument is made.
Keeping the finite and infinite claims separate is important. A formula for \(S_N\) can show how partial sums behave as \(N\) changes, but a claim about the infinite series requires a conclusion about the limit of the sequence \((S_N)\). Likewise, summation by parts is valid for every fixed \(N\); using it to prove a limit statement requires controlling the endpoint term and the remaining finite sum as \(N\) grows.
For nonnegative terms, increasing partial sums provide additional order structure. If they are bounded, they converge; if they are unbounded, they eventually exceed every fixed real number. For terms of mixed sign, neither conclusion follows from boundedness alone. The next tutorial develops convergence of a series more systematically, using the partial sums as the central object.
Check Your Understanding
Use the identities and results in this tutorial to answer each question.
- If \(S_0=0\) and \(a_n=S_n-S_{n-1}\), why does the \(N\)th partial sum equal \(S_N\)?
- Express the block sum \(\sum_{n=M+1}^{N}a_n\) in terms of partial sums, and explain why the formula also holds when \(M=0\).
- In the finite summation-by-parts identity, what are \(A_k\) and \(A_N\)?
- Why are the partial sums nondecreasing when every term is nonnegative?
- What additional condition, besides nonnegative terms, ensures that the partial sums converge to a finite real number?
- Why does a finite summation-by-parts identity by itself not establish convergence of an infinite series?