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Series and Partial Sums · Tutorial 553 of 998

Convergence Through the Partial-Sum Sequence

Use the limit of an explicitly given partial-sum sequence to decide whether a series converges and, when it does, find its value.

Beginner 9 min read

What You'll Learn

  • State series convergence precisely in terms of the partial-sum sequence.
  • Find a series value by evaluating the limit of its explicit partial-sum formula.
  • Use bounds to justify finite limits.
  • Distinguish oscillations that shrink from oscillations that persist.
  • Use subsequences to prove that a partial-sum sequence does not converge.

Why Partial Sums Decide Convergence

An infinite series is not evaluated by adding infinitely many terms all at once. Instead, its partial sums form a sequence, and convergence of the series is determined by whether that sequence approaches a finite limit.

This makes an explicitly given partial-sum formula especially useful. If the formula lets us find the limit of the partial sums, then it tells us directly whether the series converges and, if it does, what its value is. The important object is the entire sequence of partial sums—not whether individual partial sums happen to be large, small, or alternating at a few selected indices.

Definition: Let \(a_1,a_2,a_3,\dots\) be the terms of a series, and let \(S_n=a_1+a_2+\cdots+a_n\) be its \(n\)th partial sum. The series \(\sum_{k=1}^{\infty}a_k\) converges to \(L\) if the sequence of partial sums converges to the finite number \(L\): $$ \lim_{n\to\infty} S_n=L. $$ In that case, \(L\) is the value of the series. If the partial-sum sequence has no finite limit, the series diverges.

Thus, when a problem gives an explicit formula for \(S_n\), we can study that formula as a sequence. We do not need to add more and more terms numerically if the limit can be found directly.

From a Partial-Sum Limit to a Series Value

The connection between the series and its partial-sum sequence is exact: it is the definition of convergence for an infinite series. A finite limit of the partial sums is both necessary and sufficient for convergence. There is no additional condition to check once the given sequence really is the sequence of partial sums.

Here is why the limit gives the value. For each positive integer \(n\), \(S_n\) is the sum of the first \(n\) terms. Saying that \(S_n\) approaches \(L\) means that, for every chosen tolerance \(\varepsilon>0\), all sufficiently late partial sums differ from \(L\) by less than \(\varepsilon\). That is precisely what it means for the finite partial sums to settle toward the value of the infinite series.

$$ \forall \varepsilon>0,\ \exists N\in\mathbb{N} \text{ such that }n\geq N\implies |S_n-L|<\varepsilon. $$

This also explains why examining a few early partial sums is not enough. Convergence is about what happens for every sufficiently large index, not about a finite initial segment.

Worked Example: A Partial-Sum Sequence with Limit 6

Worked Example: A Partial-Sum Sequence with Limit 6

Suppose the partial sums of a series are given, for \(n\geq 1\), by

$$ S_n=6-\frac{4}{n+2}. $$

Determine whether the series converges and, if it does, find its value.

As \(n\) increases, the denominator \(n+2\) increases without bound, so the fraction approaches zero. Therefore the candidate limit is

$$ \lim_{n\to\infty}S_n = \lim_{n\to\infty}\left(6-\frac{4}{n+2}\right) = 6-0 = 6. $$

To verify the limit directly, let \(\varepsilon>0\). We have

$$ |S_n-6| = \left|-\frac{4}{n+2}\right| = \frac{4}{n+2}. $$

Choose a positive integer \(N\) large enough that \(N+2>4/\varepsilon\). For every \(n\geq N\), \(n+2\geq N+2>4/\varepsilon\), and hence \(4/(n+2)<\varepsilon\). Thus \(S_n\to 6\). The series converges, and its value is \(6\).

Check: The difference between the displayed partial sum and the proposed value is exactly \(4/(n+2)\), which tends to zero. This confirms the limit used to identify the series value.

Oscillation Does Not Always Prevent Convergence

A sequence can alternate above and below a number and still converge, provided the size of the deviations shrinks toward zero. The word “oscillates” by itself does not decide convergence. We must check whether the values settle toward a single finite limit.

Worked Example: Shrinking Oscillations with Limit 2

Suppose the partial sums are

$$ S_n=2+\frac{(-1)^n}{n},\qquad n\geq 1. $$

The factor \((-1)^n\) alternates between \(1\) and \(-1\), so the partial sums alternate around \(2\). But the size of the added term is \(1/n\), which tends to zero. In detail,

$$ |S_n-2| = \left|\frac{(-1)^n}{n}\right| = \frac{1}{n}. $$

Given any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For every \(n\geq N\),

$$ |S_n-2|=\frac{1}{n}\leq\frac{1}{N}<\varepsilon. $$

Therefore \(S_n\to 2\). The series converges to \(2\), despite the alternating partial sums.

Check: For even \(n\), \(S_n=2+1/n\); for odd \(n\), \(S_n=2-1/n\). Both values approach \(2\) as their indices grow, in agreement with the bound above.

This example gives an important distinction: persistent oscillation can prevent a limit, but oscillation whose size decreases to zero need not. The limit depends on the eventual distance of every partial sum from the proposed value.

When Partial Sums Keep Oscillating

One reliable way to show that a sequence does not converge is to find two subsequences that approach different limits. If the full sequence had a limit, every subsequence would have to approach that same limit. Therefore two different subsequential limits rule out convergence.

Theorem: If a sequence converges to \(L\), then every subsequence of it also converges to \(L\). Consequently, if two subsequences have different limits, the original sequence does not converge.

The reason for the theorem follows directly from the definition of a sequence limit. Given \(\varepsilon>0\), all terms of the original sequence from some index \(N\) onward are within \(\varepsilon\) of \(L\). A subsequence whose indices increase without bound eventually uses only indices at least \(N\), so its terms are within the same \(\varepsilon\) of \(L\). Thus the subsequence has limit \(L\) as well.

Worked Example: Persistent Alternation and Divergence

Suppose the partial sums are

$$ S_n=(-1)^n,\qquad n\geq 1. $$

At every even index \(n=2k\), where \(k\geq 1\),

$$ S_{2k}=(-1)^{2k}=1. $$

This even-indexed subsequence is constantly \(1\), so it converges to \(1\). At every odd index \(n=2k-1\),

$$ S_{2k-1}=(-1)^{2k-1}=-1. $$

This odd-indexed subsequence is constantly \(-1\), so it converges to \(-1\). Since \(1\neq -1\), the two subsequences have different limits. The full partial-sum sequence cannot converge. Therefore the series diverges.

Check: The values of the partial sums are \(-1,1,-1,1,\dots\). The even and odd subsequences confirm that this alternation does not shrink toward one finite value.

Notice that this is a proof of divergence, not just an observation that some terms differ. A sequence can have unequal consecutive terms and still converge. Here, the persistent separation between the two subsequences is what proves that no single limit is possible.

Unbounded Partial Sums Also Mean Divergence

A sequence that grows without bound cannot converge to a finite real number. For example, if the partial-sum formula is \(S_n=n\), then \(S_n\) becomes arbitrarily large as \(n\) increases. It therefore has no finite limit, and the corresponding series diverges.

More generally, if an explicit formula shows that \(|S_n|\) becomes arbitrarily large, that is enough to rule out a finite limit. But boundedness alone is not enough to prove convergence: a bounded sequence can continue to oscillate, as the sequence \(S_n=(-1)^n\) does.

Key takeaway: To decide convergence from a partial-sum formula, determine whether the full sequence \(S_n\) approaches one finite number. A finite limit gives the series value; unbounded growth or persistent oscillation rules out convergence.

A Common Mistake: Treating Every Alternation as Divergence

A tempting but incorrect argument is: “The partial sums alternate, so the series diverges.” Alternation alone is not decisive. Consider the partial sums in the shrinking-oscillation example. They alternate, but their distance from \(2\) is \(1/n\), and that distance tends to zero.

For instance, the following calculation would be wrong:

$$ S_n=2+\frac{(-1)^n}{n} \quad\Longrightarrow\quad \text{the values alternate} \quad\Longrightarrow\quad \text{the series diverges}. $$

The correction is to measure the size of the oscillation:

$$ |S_n-2|=\frac{1}{n}\longrightarrow 0. $$

Therefore these partial sums converge to \(2\). Compare this with \(S_n=(-1)^n\): its even and odd subsequences remain at \(1\) and \(-1\), so the oscillation does not shrink and the sequence diverges. Always test the limit of the partial sums rather than relying on the appearance of alternation.

A Practical Method

When the partial-sum sequence is given explicitly, a short, careful process keeps the conclusion tied to the definition.

1
Identify the sequence.
Write down the given formula for \(S_n\) and the indices for which it is defined.
2
Find its limiting behavior.
Evaluate the limit if one is suggested by the formula; if oscillation occurs, check whether its size shrinks.
3
Justify the conclusion.
For convergence, show that the partial sums approach a finite number. For divergence, exhibit unbounded growth or incompatible subsequential limits.
4
State the series result.
A finite limit is the series value. If there is no finite limit, the series diverges.

This method applies to explicit partial-sum sequences whether their formulas involve fractions, powers, or alternating signs. It does not require guessing a pattern from a list of numerical partial sums. The decisive evidence is the limit behavior of the entire sequence.

Check Your Understanding

For each question, use the behavior of the partial-sum sequence to decide what can be concluded about the series.

  1. If \(S_n=4+\frac{3}{n}\), does the series converge? If so, what is its value?
  2. If \(S_n=1+\frac{(-1)^n}{n+2}\), does alternation prevent the partial sums from converging? Explain using their distance from \(1\).
  3. If \(S_{2k}\to 5\) and \(S_{2k-1}\to 2\), can the full partial-sum sequence converge? Why or why not?
  4. What information about an explicit partial-sum sequence is enough to identify the value of a convergent series?
  5. Why does boundedness of the partial sums, by itself, not establish convergence?

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