Infinite Series as Limits of Partial Sums

From Finite Sums to Infinite Series

An infinite series begins with a sequence of terms, but its meaning depends on more than the terms considered one at a time. To define what the terms add up to, we form longer and longer finite sums and ask whether those sums approach a finite limit.

This distinction matters in computation and in reasoning. A finite sum uses a specified number of terms. An infinite series is defined by the limiting behavior of all its partial sums. A finite partial sum may approximate the series value, but it is not automatically equal to that value.

Partial Sums and the Definition of a Series

Let \(a_1,a_2,a_3,\ldots\) be a sequence of real numbers. For each positive integer \(N\), its \(N\)th partial sum is the finite sum of its first \(N\) terms:

S_N=\sum_{n=1}^{N}a_n=a_1+a_2+\cdots+a_N.

The sequence \(S_1,S_2,S_3,\ldots\) is called the sequence of partial sums. An infinite series is defined by asking whether this sequence has a finite limit as \(N\) tends to infinity.

Definition: The infinite series \(\sum_{n=1}^{\infty}a_n\) converges if the sequence of partial sums \(S_N=\sum_{n=1}^{N}a_n\) has a finite limit as \(N\to\infty\). If that limit is \(S\), then \(S\) is the value of the series, and we write \(\sum_{n=1}^{\infty}a_n=S\). If the partial sums do not have a finite limit, the series diverges.

The symbol \(\sum_{n=1}^{\infty}a_n\) indicates an infinite series. Its value is defined only when the partial sums converge. By contrast, \(\sum_{n=1}^{N}a_n\) is an ordinary finite sum for a particular \(N\); it is the partial sum \(S_N\), whether or not the infinite series converges.

Why the Limit Defines the Infinite Sum

The definition says that a convergent infinite series is not obtained by carrying out an endless addition in a final step. Instead, its value describes what the finite partial sums approach. The limit condition makes the idea of “approach” precise.

Meaning of the limit: If \(S_N\to S\), then for every \(\varepsilon>0\), there is a positive integer \(N_0\) such that \(N\geq N_0\) implies \(|S_N-S|<\varepsilon\).

In other words, once enough terms have been included, every later partial sum is within any prescribed positive distance \(\varepsilon\) of \(S\). This is why a partial sum can serve as a finite approximation to the series value: for any chosen accuracy, convergence guarantees that sufficiently many terms produce an approximation within that accuracy.

Choose a finite number of terms.
For a positive integer \(N\), calculate \(S_N=a_1+\cdots+a_N\).

View the partial sums as a sequence.
As \(N\) increases, the values \(S_N\) may approach a finite number, grow without bound, or fail to settle toward one value.

Use the limit to define the series value.
If \(S_N\to S\), the series converges to \(S\). Otherwise, it diverges and has no finite value under this definition.

The definition also explains why the infinite-series notation should not be read as “add a certain large but finite number of terms.” Any specific choice of \(N\) produces only \(S_N\). The infinite series concerns the behavior of the entire sequence of these finite sums.

Worked Examples

Worked Example: A Series with Only Three Nonzero Terms

Define \(a_1=3\), \(a_2=-2\), \(a_3=5\), and \(a_n=0\) for every \(n\geq4\). Find the value of \(\sum_{n=1}^{\infty}a_n\), if it exists.

First calculate the partial sums:

S_1=3,\qquad S_2=3+(-2)=1,\qquad S_3=3+(-2)+5=6.

For every \(N\geq4\), all additional terms are zero, so

S_N=3+(-2)+5+\underbrace{0+\cdots+0}_{N-3\text{ terms}}=6.

Thus \(S_N=6\) for every \(N\geq3\), and therefore \(\lim_{N\to\infty}S_N=6\). The series converges to \(6\). To check, every partial sum from \(S_3\) onward is exactly \(6\), so the partial sums do approach the stated value.

Worked Example: Alternating Terms That Do Not Produce a Limit

Consider \(a_n=(-1)^{n+1}\), whose terms are \(1,-1,1,-1,\ldots\). Determine whether its infinite series converges.

The first partial sums are

S_1=1,\qquad S_2=1-1=0,\qquad S_3=1-1+1=1,\qquad S_4=1-1+1-1=0.

In general, for each positive integer \(k\), \(S_{2k}=0\) and \(S_{2k-1}=1\). If the sequence of partial sums had a limit \(L\), then both its even-indexed and odd-indexed subsequences would also approach \(L\). But the even-indexed partial sums are always \(0\), while the odd-indexed partial sums are always \(1\); they cannot approach the same number. Hence the partial sums have no limit, and the series diverges.

As a direct check, partial sums continue to alternate between \(0\) and \(1\), so they do not settle near a single finite value.

Worked Example: A Finite Partial Sum Is Not the Infinite Series

Consider the harmonic series \(\sum_{n=1}^{\infty}\frac1n\). Calculate its fourth partial sum, then determine whether the infinite series has a finite value.

The fourth partial sum is a finite sum:

S_4=1+\frac12+\frac13+\frac14 =\frac{12}{12}+\frac{6}{12}+\frac{4}{12}+\frac{3}{12} =\frac{25}{12}.

This computes \(S_4\), not the value of the infinite series. To decide whether a finite series value exists, consider larger partial sums. For each positive integer \(m\), group the terms from \(2^{j-1}+1\) to \(2^j\), for \(j=1,\ldots,m\). In the \(j\)th group there are \(2^{j-1}\) terms, and each is at least \(1/2^j\). Therefore that group contributes at least \(2^{j-1}/2^j=1/2\). It follows that

S_{2^m} =1+\sum_{j=1}^{m}\sum_{n=2^{j-1}+1}^{2^j}\frac1n \geq 1+\sum_{j=1}^{m}\frac12 =1+\frac{m}{2}.

As \(m\) increases, \(1+m/2\) grows without bound. Thus the partial sums \(S_N\) are unbounded: among them are the values \(S_{2^m}\), which exceed any fixed bound when \(m\) is large enough. A convergent sequence must be bounded, so these partial sums cannot converge to a finite limit. The series diverges. This check confirms that \(25/12\) is just the fourth partial sum, not a finite value for the infinite series.

A Common Mistake: Treating an Approximation as the Sum

A common error is to calculate a few terms and then label the result as the value of the infinite series. For instance, the calculation

\sum_{n=1}^{4}\frac1n=\frac{25}{12}

is correct, but it states only that the fourth partial sum is \(25/12\). It does not establish that \(\sum_{n=1}^{\infty}\frac1n=25/12\). In fact, as the preceding example shows, the harmonic series diverges.

The correction is to name the finite calculation accurately and then address the limit separately: first write \(S_4=25/12\); then determine whether \(S_N\) has a finite limit as \(N\to\infty\). A finite calculation alone cannot settle that limit.

Why This Definition Matters

Defining series through partial sums gives a single framework for different behaviors. Partial sums may eventually become constant, may fail to settle, or may grow without bound. In each case, the question is the same: does the sequence \(S_N\) approach a finite limit?

This framework also keeps exact values separate from approximations. When a series converges to \(S\), a partial sum \(S_N\) can approximate \(S\), but it is generally a different number. When a series diverges, no finite number can be its value under the definition, even though every individual partial sum is finite.

Key takeaway: An infinite series is defined by the limit of its sequence of finite partial sums. The finite sum \(S_N\) is one approximation in that sequence; it is not, by itself, the infinite series value.

Check Your Understanding

Answer these questions using the definition of an infinite series.

For a sequence of terms \(a_n\), how is its \(N\)th partial sum \(S_N\) defined?
What condition on the partial sums makes \(\sum_{n=1}^{\infty}a_n\) converge?
If \(S_5=12\), what does this tell you, and what does it not establish about the infinite series?
Why does the alternating series with terms \(1,-1,1,-1,\ldots\) diverge under the partial-sum definition?
Can an infinite series have a finite value if its partial sums are unbounded? Explain using the definition.