Unit 7, lesson 93

Series and Partial Sums

What it means to add infinitely many numbers, how partial sums turn that question into a limit of a sequence, and how the terms, the sum and the remainder differ.

  • Theory part 185
  • Practice part 186
  • Infinite Series

Walk halfway to a wall, then half the remaining distance, then half of what is left, and keep going. You take infinitely many steps, yet you never pass the wall, and the total distance is exactly the distance to the wall. Infinite sums like this one need a careful meaning, because you can never actually finish adding. The idea is to add a finite number of terms at a time and ask where those running totals are headed.

Key terms

Infinite series

Definition Given a sequence of real numbers (an)n≥1(a_n)_{n \ge 1}, the infinite series with terms ana_n is the formal expression

∑n=1∞an=a1+a2+a3+⋯ .\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots.

The index may start at any integer, for example ∑n=0∞an\sum_{n=0}^{\infty} a_n, and the number ana_n is called the nnth term of the series.

In plain words A series is a list of numbers that you intend to add up, one after another, forever. Writing it down does not yet say whether the total exists.

Example The walk to the wall, measured in units of the starting distance, is the series

∑n=1∞12n=12+14+18+⋯ .\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots.

Partial sums

Definition The NNth partial sum of ∑n=1∞an\sum_{n=1}^{\infty} a_n is the finite sum

sN=∑n=1Nan=a1+a2+⋯+aN.s_N = \sum_{n=1}^{N} a_n = a_1 + a_2 + \cdots + a_N.

The partial sums form a new sequence (sN)N≥1(s_N)_{N \ge 1}, and they satisfy sN=sN−1+aNs_N = s_{N-1} + a_N for N≥2N \ge 2.

In plain words A partial sum is the running total after you have added the first NN terms.

Example For ∑12n\sum \frac{1}{2^n} the running totals are

NNsNs_N
110.50.5
220.750.75
330.8750.875
440.93750.9375
550.968750.96875

Each time the gap to 11 is cut in half, and in fact sN=1−12Ns_N = 1 - \frac{1}{2^N}.

Convergence through partial sums

Definition The series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges if its sequence of partial sums converges to a real number SS, that is

lim⁡N→∞sN=S.\lim_{N \to \infty} s_N = S.

Then SS is called the sum of the series and we write ∑n=1∞an=S\sum_{n=1}^{\infty} a_n = S.

In plain words An infinite sum is the limit of the running totals. If the running totals settle on one number, that number is the sum.

Example Suppose the partial sums of a series are sN=2NN+1s_N = \frac{2N}{N + 1}. Then sN→2s_N \to 2, so the series converges with sum 22. You can even recover the terms. We get a1=s1=1a_1 = s_1 = 1, and for n≥2n \ge 2

an=sn−sn−1=2nn+1−2(n−1)n=2n(n+1).a_n = s_n - s_{n-1} = \frac{2n}{n + 1} - \frac{2(n - 1)}{n} = \frac{2}{n(n + 1)}.

The formula also gives a1=1a_1 = 1, so

∑n=1∞2n(n+1)=2.\sum_{n=1}^{\infty} \frac{2}{n(n + 1)} = 2.

Divergence

Definition A series diverges if its sequence of partial sums does not converge to a real number. This includes sN→∞s_N \to \infty, sN→−∞s_N \to -\infty, and partial sums that oscillate without settling.

In plain words If the running totals never close in on a single finite number, the series has no sum.

Example For ∑n=1∞n\sum_{n=1}^{\infty} n the partial sums are sN=N(N+1)2→∞s_N = \frac{N(N + 1)}{2} \to \infty. For ∑n=1∞(−1)n+1=1−1+1−1+⋯\sum_{n=1}^{\infty} (-1)^{n+1} = 1 - 1 + 1 - 1 + \cdots the partial sums are 1,0,1,0,…1, 0, 1, 0, \ldots, which oscillate. Both series diverge, in two different ways.

The terms versus the sum

Definition A series carries two sequences, the sequence of terms (an)(a_n) and the sequence of partial sums (sN)(s_N). Convergence of the series means convergence of (sN)(s_N). Convergence of (an)(a_n) is a different question, and lim⁡an\lim a_n is not the sum of the series.

In plain words Do not mix up the numbers you are adding with the total you get. The terms can settle down while the total runs off to infinity.

Example Let an=nn+1a_n = \frac{n}{n + 1}. The sequence of terms converges, since an→1a_n \to 1. But every term is at least 12\frac{1}{2}, so sN≥N2→∞s_N \ge \frac{N}{2} \to \infty and the series ∑nn+1\sum \frac{n}{n + 1} diverges.

Remainder

Definition If ∑n=1∞an\sum_{n=1}^{\infty} a_n converges with sum SS, the remainder after NN terms is

RN=S−sN=∑n=N+1∞an.R_N = S - s_N = \sum_{n=N+1}^{\infty} a_n.

Because sN→Ss_N \to S, the remainder satisfies RN→0R_N \to 0.

In plain words The remainder is the part of the sum you leave out when you stop after NN terms. It is exactly the error of using sNs_N in place of SS.

Example For ∑12n\sum \frac{1}{2^n} with S=1S = 1, the remainder is

RN=1−(1−12N)=12N.R_N = 1 - \left(1 - \frac{1}{2^N}\right) = \frac{1}{2^N}.

After 1010 terms the error is 11024\frac{1}{1024}, a little under 0.0010.001.

Check yourself

  1. The partial sums of a series are sN=3−1Ns_N = 3 - \frac{1}{N}. Find the sum and a formula for the terms.
Show the answer

The sum is lim⁡sN=3\lim s_N = 3. The first term is a1=s1=2a_1 = s_1 = 2, and for n≥2n \ge 2 the terms are

an=sn−sn−1=1n−1−1n=1n(n−1).a_n = s_n - s_{n-1} = \frac{1}{n - 1} - \frac{1}{n} = \frac{1}{n(n - 1)}.
  1. Does ∑n=1∞1\sum_{n=1}^{\infty} 1 converge?
Show the answer

No. The partial sums are sN=Ns_N = N, which grow without bound, so the series diverges.

  1. A convergent series has sum 55 and s20=4.9s_{20} = 4.9. What is R20R_{20}?
Show the answer

The remainder is the sum minus the partial sum, so R20=S−s20=5−4.9=0.1R_{20} = S - s_{20} = 5 - 4.9 = 0.1.

How this lesson connects