Unit 10: Infinite Sequences and Series
- Syllabus
- 2020
- Section
- —
- Level
- —

For the infinite series ∑k=1∞ak, the nth partial sum Sn adds only the first n terms. The infinite series converges to S exactly when the sequence S1,S2,S3,… approaches the finite real number S.
S_n=\sum_{k=1}^{n}a_k,\qquad \sum_{k=1}^{\infty}a_k=S\iff\lim_{n\to\infty}S_n=S
Example: since 1/[k(k+1)]=1/k−1/(k+1),
Sn=k=1∑nk(k+1)1=(1−21)+(21−31)+⋯+(n1−n+11)=1−n+11.
Therefore limn→∞Sn=1, so ∑k=1∞1/[k(k+1)] converges and its sum is 1. By contrast, for ∑k=1∞1, Sn=n→∞, so the series diverges.
A convergent series still contains infinitely many terms; convergence does not mean the adding stops. It means the finite partial sums approach one finite value. Writing that a divergent series has a sum of ∞ is not the same as convergence to a real number.
A geometric series has the form a+ar+ar2+⋯, so dividing any term after the first by its preceding nonzero term gives the same common ratio r. For a nonzero series, its terms shrink toward zero exactly when ∣r∣<1.
\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad\text{when }|r|<1;\qquad |r|\ge1\Rightarrow\text{divergence}
The finite partial sum through exponent N is SN=a(1−rN+1)/(1−r) for r=1. When ∣r∣<1, rN+1→0, leaving a/(1−r). If ∣r∣≥1, that power does not approach zero, so the partial sums do not approach a finite value.
Example: 6−3+23−43+⋯ is geometric because each term is the previous term multiplied by −1/2. Here a=6 and r=−1/2. Since ∣−1/2∣<1, the series converges, and
S=1−(−1/2)6=3/26=4.
The alternating signs do not prevent convergence because the term magnitudes shrink.
Use the actual first term of the displayed series as a. For example, ∑n=1∞3(1/2)n begins with 3/2, not 3; either reindex from zero or use a=3/2. Also, r=−1 alternates but does not shrink, so it diverges.
If ∑an converges, adding the next term must eventually make almost no change to its partial sum. Because an=Sn−Sn−1, convergence of the partial sums forces an→0. Therefore any failure of the terms to approach zero proves divergence.
\lim_{n\to\infty}a_n\ne0\text{ or does not exist}\Rightarrow\sum_{n=1}^{\infty}a_n\text{ diverges};\qquad \lim_{n\to\infty}a_n=0\Rightarrow\text{inconclusive}
Example 1: for ∑n=1∞n/(n+1),
n→∞limn+1n=1=0.
The terms keep contributing about 1, so the nth term test proves that the series diverges. No other convergence test is needed.
Example 2: for ∑n=1∞1/n, the term limit is 0. The nth term test is inconclusive: it does not say the series converges. A different test is required to determine its behavior.
The test is designed to prove divergence, not convergence. The statement “a convergent series has an→0” is true, but its converse is false: terms approaching zero is necessary, not sufficient, for the accumulated sum to approach a finite value.
Suppose an=f(n) and, for all x≥N, the function f is continuous, positive, and decreasing. Then the terms can be compared with adjacent strips under the graph, so the infinite series and corresponding improper integral either both converge or both diverge.
\sum_{n=N}^{\infty}a_n\text{ converges}\iff\int_N^{\infty}f(x),dx\text{ converges},\qquad a_n=f(n)
Example: for ∑n=1∞1/(n2+1), let f(x)=1/(x2+1). For x≥1, f is continuous and positive, and f′(x)=−2x/(x2+1)2<0, so it is decreasing. Also,
∫1∞x2+1dx=b→∞lim[arctanx]1b=2π−4π=4π.
The improper integral converges, so the series converges by the integral test.
The value π/4 in the example is the integral's value, not the sum of the series. The integral test determines convergence behavior; it does not normally calculate the exact series sum. If positivity or decreasing behavior fails, this test has not been justified.