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Sequence of Partial Sums

Sequence of Partial Sums

The sequence of nth partial sums of a series.

 

Example showing partial sums S1=1, S2=1+1/4, S3=1+1/4+1/9, S4=1+1/4+1/9+1/16 for series 1+1/4+1/9+1/16+…+1/n²+…

 

 

See also

Convergence tests, infinite series

Key Formula

{Sn}={S1,S2,S3,}whereSn=k=1nak=a1+a2+a3++an\{S_n\} = \{S_1,\, S_2,\, S_3,\, \ldots\} \quad \text{where} \quad S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n
Where:
  • SnS_n = The nth partial sum — the sum of the first n terms of the series
  • aka_k = The kth term of the original series
  • nn = The number of terms being summed (a positive integer)

Worked Example

Problem: Find the first five terms of the sequence of partial sums for the series k=112k\sum_{k=1}^{\infty} \frac{1}{2^k}.
Step 1: Identify the terms of the series. The kth term is ak=12ka_k = \frac{1}{2^k}, so the first several terms are 12,14,18,116,132,\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \ldots
a1=12,a2=14,a3=18,a4=116,a5=132a_1 = \tfrac{1}{2},\quad a_2 = \tfrac{1}{4},\quad a_3 = \tfrac{1}{8},\quad a_4 = \tfrac{1}{16},\quad a_5 = \tfrac{1}{32}
Step 2: Compute S1S_1, the first partial sum.
S1=a1=12S_1 = a_1 = \frac{1}{2}
Step 3: Compute S2S_2 and S3S_3 by adding the next term each time.
S2=12+14=34,S3=34+18=78S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}, \qquad S_3 = \frac{3}{4} + \frac{1}{8} = \frac{7}{8}
Step 4: Continue for S4S_4 and S5S_5.
S4=78+116=1516,S5=1516+132=3132S_4 = \frac{7}{8} + \frac{1}{16} = \frac{15}{16}, \qquad S_5 = \frac{15}{16} + \frac{1}{32} = \frac{31}{32}
Step 5: Write the sequence of partial sums. Notice each term follows the pattern Sn=112nS_n = 1 - \frac{1}{2^n}, and as nn \to \infty, Sn1S_n \to 1. So the series converges to 1.
{Sn}={12,  34,  78,  1516,  3132,  }\{S_n\} = \left\{\frac{1}{2},\; \frac{3}{4},\; \frac{7}{8},\; \frac{15}{16},\; \frac{31}{32},\; \ldots \right\}
Answer: The first five partial sums are 12,34,78,1516,3132\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \frac{31}{32}. The sequence converges to 1.

Another Example

This example shows a telescoping series where you can find a closed-form formula for SnS_n and then take its limit, rather than just listing numerical values of partial sums.

Problem: Find a closed-form expression for the nth partial sum of the telescoping series k=1(1k1k+1)\sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) and determine whether the series converges.
Step 1: Write out the partial sum SnS_n by expanding the first several terms.
Sn=(1112)+(1213)+(1314)++(1n1n+1)S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)
Step 2: Most terms cancel in pairs (this is why it is called a telescoping series). After cancellation, only the first and last parts remain.
Sn=11n+1S_n = 1 - \frac{1}{n+1}
Step 3: Check a few values: S1=112=12S_1 = 1 - \frac{1}{2} = \frac{1}{2}, S2=113=23S_2 = 1 - \frac{1}{3} = \frac{2}{3}, S3=114=34S_3 = 1 - \frac{1}{4} = \frac{3}{4}. These match direct computation.
{Sn}={12,  23,  34,  45,  }\{S_n\} = \left\{\frac{1}{2},\; \frac{2}{3},\; \frac{3}{4},\; \frac{4}{5},\; \ldots\right\}
Step 4: Take the limit of the sequence of partial sums to determine convergence.
limnSn=limn(11n+1)=1\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1
Answer: The nth partial sum is Sn=11n+1S_n = 1 - \frac{1}{n+1}, and the series converges to 1.

Frequently Asked Questions

What is the difference between a series and a sequence of partial sums?
A series is the expression k=1ak\sum_{k=1}^{\infty} a_k, which represents the idea of adding infinitely many terms. The sequence of partial sums {S1,S2,S3,}\{S_1, S_2, S_3, \ldots\} is the actual sequence you build by computing finite sums one term at a time. The series converges if and only if this sequence of partial sums converges to a finite limit.
How do you tell if a series converges using partial sums?
Compute (or find a formula for) the nth partial sum SnS_n, then evaluate limnSn\lim_{n \to \infty} S_n. If this limit exists and is a finite number LL, the series converges to LL. If the limit does not exist or is infinite, the series diverges.
Can you always find a formula for the nth partial sum?
No. Closed-form expressions for SnS_n exist for certain types of series such as geometric series, telescoping series, and a few others. For most series, you cannot write SnS_n in a simple closed form, which is why convergence tests (like the ratio test or comparison test) are so important — they let you determine convergence without needing an explicit formula for SnS_n.

Sequence of Partial Sums vs. nth Partial Sum

Sequence of Partial Sumsnth Partial Sum
What it isThe entire sequence {S1,S2,S3,}\{S_1, S_2, S_3, \ldots\}A single value SnS_n for one particular nn
Formula{Sn}n=1\{S_n\}_{n=1}^{\infty} where each Sn=k=1nakS_n = \sum_{k=1}^{n} a_kSn=a1+a2++anS_n = a_1 + a_2 + \cdots + a_n
PurposeAnalyze convergence by studying the behavior of the whole sequenceApproximate the sum of a series using the first nn terms
Example output{12,34,78,}\{\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \ldots\} (a list of values)S3=78S_3 = \frac{7}{8} (a single number)

Why It Matters

The sequence of partial sums is the fundamental tool for defining what it means for an infinite series to converge. In calculus courses, you encounter it whenever you study geometric series, telescoping series, or power series representations of functions. Understanding this concept bridges the gap between finite addition (which you already know) and the idea of summing infinitely many terms.

Common Mistakes

Mistake: Confusing the original sequence {an}\{a_n\} with the sequence of partial sums {Sn}\{S_n\}.
Correction: The terms ana_n are the individual pieces being added. The partial sums SnS_n are cumulative totals. For example, if an=12na_n = \frac{1}{2^n}, then a3=18a_3 = \frac{1}{8} but S3=78S_3 = \frac{7}{8}. These are very different sequences.
Mistake: Assuming that if an0a_n \to 0 then SnS_n must converge to a finite limit.
Correction: The terms approaching zero is necessary but not sufficient for convergence. The harmonic series 1k\sum \frac{1}{k} has ak0a_k \to 0, yet SnS_n \to \infty. You must examine the limit of SnS_n itself (or use a convergence test) to determine whether the series converges.

Related Terms

  • SequenceA sequence of partial sums is itself a sequence
  • nth Partial SumEach individual term in the sequence of partial sums
  • SeriesThe sum whose convergence is determined by partial sums
  • Convergence TestsMethods used when closed-form partial sums are unavailable
  • Infinite SeriesConverges when its sequence of partial sums has a finite limit
  • Geometric SeriesA classic series with a known partial sum formula
  • Telescoping SeriesSeries where partial sums simplify by cancellation