Infinite Series & Convergence Tests
Sections 5.2–5.3: partial sums, telescoping & geometric series, divergence test, p-series, integral test
A sequence is a list of numbers \(a_1, a_2, a_3, \dots\). A series is what you get when you try to add up that list:
\[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots \]
You can’t literally add infinitely many numbers, so we define the sum through a sequence we can compute. That idea, the sequence of partial sums, drives everything this week.
Part 1: Series and partial sums (5.2)
Partial sums
The \(N\)th partial sum is the sum of the first \(N\) terms:
\[ S_N = \sum_{n=1}^{N} a_n = a_1 + a_2 + \cdots + a_N \]
The series \(\sum a_n\) converges to \(S\) if the sequence of partial sums converges: \(\displaystyle\lim_{N\to\infty} S_N = S\). Otherwise the series diverges.
Keep the two sequences separate in your head: \(\{a_n\}\) is the list of terms, and \(\{S_N\}\) is the list of running totals.
Example. First four partial sums of \(\displaystyle\sum_{n=1}^\infty \frac{1}{n}\):
\[ S_1 = 1, \quad S_2 = 1 + \tfrac12 = \tfrac32, \quad S_3 = \tfrac32 + \tfrac13 = \tfrac{11}{6}, \quad S_4 = \tfrac{11}{6} + \tfrac14 = \tfrac{25}{12} \]
Example. First four partial sums of \(\displaystyle\sum_{n=1}^\infty 3\cos(n\pi)\). Since \(\cos(n\pi) = (-1)^n\), the terms are \(-3, 3, -3, 3, \dots\):
\[ S_1 = -3, \quad S_2 = 0, \quad S_3 = -3, \quad S_4 = 0 \]
The partial sums bounce between \(-3\) and \(0\) forever, so \(\lim S_N\) does not exist and the series diverges.
When WeBWorK says “enter exact values,” type fractions like 25/12, not 2.0833. For trig terms, evaluate the trig function first (\(\sin(\pi/2) = 1\), \(\cos(\pi) = -1\), and so on).
Practice: first four partial sums of \(\displaystyle\sum_{n=1}^\infty \frac{4}{n}\) and of \(\displaystyle\sum_{n=1}^\infty 2\sin\!\left(\frac{n\pi}{2}\right)\).
Writing a series in sigma notation (and re-indexing)
The same series can be written with the index starting anywhere. Shifting where the index starts means shifting the formula to match.
Example. \(\displaystyle 5 + \frac52 + \frac53 + \frac54 + \cdots\)
\[ \sum_{n=0}^{\infty} \frac{5}{n+1} \qquad\text{or}\qquad \sum_{n=1}^{\infty} \frac{5}{n} \]
Check: plug in the first index value and confirm you get the first term.
Example. \(7 - 7 + 7 - 7 + \cdots\). Alternating signs come from a power of \(-1\):
\[ \sum_{n=0}^{\infty} 7(-1)^n \qquad\text{or}\qquad \sum_{n=1}^{\infty} 7(-1)^{n+1} \]
(Starting at \(n=1\), we need the first term positive, so the exponent must be even when \(n=1\): \((-1)^{n+1}\) or \((-1)^{n-1}\) both work.)
To start the index one higher, replace \(n\) with \(n-1\) everywhere in the formula. To start one lower, replace \(n\) with \(n+1\).
Going backwards: from \(S_n\) to \(a_n\)
If someone hands you a formula for the partial sums, you can recover the terms, because consecutive partial sums differ by exactly one term:
\[ a_n = S_n - S_{n-1}, \qquad a_1 = S_1 \]
Example. Suppose \(S_n = 3 - \dfrac{3}{n+1}\).
\[ a_n = \left(3 - \frac{3}{n+1}\right) - \left(3 - \frac{3}{n}\right) = \frac{3}{n} - \frac{3}{n+1} = \frac{3}{n(n+1)} \]
And since \(\displaystyle\lim_{n\to\infty} S_n = 3 - 0 = 3\), the series converges and \(\displaystyle\sum_{n=1}^\infty \frac{3}{n(n+1)} = 3\).
Example. Suppose \(S_n = 4\ln(n+1)\). Then \(a_n = 4\ln(n+1) - 4\ln(n) = 4\ln\!\left(\frac{n+1}{n}\right)\), and \(S_n \to \infty\), so the series diverges ($S = $ DNE).
The formula \(a_n = S_n - S_{n-1}\) is guaranteed only for \(n \ge 2\). It also gives the right \(a_1\) exactly when the \(S_n\) formula gives \(S_0 = 0\). In the first example \(S_0 = 3 - 3 = 0\), so we’re fine.
Practice: find \(a_n\) and \(S\) if \(S_n = 8 - \dfrac{8}{n+1}\); if \(S_n = 2\sqrt{n}\).
Using partial sums to see divergence
Example. \(\displaystyle\sum_{n=1}^\infty \frac{3n}{n+1}\). The terms are \(\tfrac32, 2, \tfrac94, \tfrac{12}{5}, \dots\), so
\[ S_1 = \tfrac32, \quad S_2 = \tfrac72, \quad S_3 = \tfrac{23}{4}, \quad S_4 = \tfrac{163}{20} \]
Each new term is close to \(3\) (because \(\frac{3n}{n+1} \to 3\)), so the running total grows by roughly \(3\) every step and \(S_N \to \infty\). The series diverges. We’ll make this reasoning official with the Divergence Test in Part 2.
Telescoping series
A series telescopes when each term has the form \(f(n) - f(n+1)\). Almost everything cancels in the partial sum:
\[ S_k = \big(f(1) - f(2)\big) + \big(f(2) - f(3)\big) + \cdots + \big(f(k) - f(k+1)\big) = f(1) - f(k+1) \]
So the series converges exactly when \(\lim_{k\to\infty} f(k+1)\) exists, and then
\[ \sum_{n=1}^\infty \big(f(n) - f(n+1)\big) = f(1) - \lim_{k\to\infty} f(k+1) \]
Example (partial fractions first). \(\displaystyle\sum_{n=1}^\infty \frac{5}{(n+2)(n+3)}\)
Partial fractions: \(\dfrac{5}{(n+2)(n+3)} = \dfrac{5}{n+2} - \dfrac{5}{n+3}\). This is \(f(n) - f(n+1)\) with \(f(n) = \dfrac{5}{n+2}\).
\[ S_k = \frac{5}{3} - \frac{5}{k+3} \quad\longrightarrow\quad \frac53 \]
Sanity check: \(S_1 = \frac{5}{12}\) directly, and the formula gives \(\frac53 - \frac54 = \frac{5}{12}\). ✓
Example (converges). \(\displaystyle\sum_{n=1}^\infty \left(\frac{1}{\sqrt n} - \frac{1}{\sqrt{n+1}}\right)\): \(\;S_k = 1 - \dfrac{1}{\sqrt{k+1}} \to 1\).
Example (diverges, oscillates). \(\displaystyle\sum_{n=1}^\infty \big(\cos(n) - \cos(n+1)\big)\): \(\;S_k = \cos(1) - \cos(k+1)\). Since \(\cos(k+1)\) keeps oscillating, \(\lim S_k\) does not exist. Diverges.
Example (diverges, log). \(\displaystyle\sum_{n=1}^\infty \ln\!\left(\frac{n}{n+1}\right)\). Use a log rule to split: \(\ln n - \ln(n+1)\), so \(S_k = \ln 1 - \ln(k+1) = -\ln(k+1) \to -\infty\). Diverges.
Look for: a rational function whose denominator factors into consecutive-ish pieces (partial fractions), a difference like \(g(n) - g(n+1)\) already written out, or a log of a quotient \(\ln\frac{n}{n+1}\).
Practice: \(\displaystyle\sum_{n=1}^\infty \frac{4}{(n+1)(n+2)}\), \(\;\displaystyle\sum_{n=1}^\infty \left(\frac{1}{n^3} - \frac{1}{(n+1)^3}\right)\), \(\;\displaystyle\sum_{n=1}^\infty \big(\arctan(n+1) - \arctan(n)\big)\) (careful with the order of subtraction)
Algebra with convergent series
If \(\sum a_n = A\) and \(\sum b_n = B\) both converge, then
\[ \sum (a_n \pm b_n) = A \pm B, \qquad \sum c\,a_n = cA \]
Changing the starting index just removes (or adds) finitely many terms:
\[ \sum_{n=2}^\infty a_n = A - a_1, \qquad \sum_{n=1}^\infty a_{n+1} = a_2 + a_3 + \cdots = A - a_1 \]
Example. Suppose \(\sum_{n=1}^\infty a_n = 3\), \(\sum_{n=1}^\infty b_n = -1\), \(a_1 = 2\), \(b_1 = 4\).
- \(\displaystyle\sum_{n=1}^\infty (2a_n - b_n) = 2(3) - (-1) = 7\)
- \(\displaystyle\sum_{n=2}^\infty (a_n + b_n) = (3 - 2) + (-1 - 4) = -4\)
- \(\displaystyle\sum_{n=1}^\infty (a_{n+1} - 2b_{n+1}) = (3-2) - 2(-1-4) = 11\)
The harmonic series
\[ \sum_{n=1}^\infty \frac1n = 1 + \frac12 + \frac13 + \frac14 + \cdots \quad \textbf{diverges} \]
even though its terms go to \(0\). One way to see it: group the terms,
\[ 1 + \frac12 + \underbrace{\left(\frac13 + \frac14\right)}_{\ge \frac12} + \underbrace{\left(\frac15 + \cdots + \frac18\right)}_{\ge \frac12} + \underbrace{\left(\frac19 + \cdots + \frac1{16}\right)}_{\ge\frac12} + \cdots \]
We keep adding at least \(\frac12\), forever.
Adding or deleting finitely many terms never changes whether a series converges. So
\[ \sum_{n=1}^\infty \frac{1}{n+10} = \frac1{11} + \frac1{12} + \cdots = \sum_{n=11}^\infty \frac1n \]
is the harmonic series with its first ten terms removed, and it still diverges.
\(\sum_{n=1}^\infty \frac{1}{n+10}\) equals \(\sum_{n=\mathbf{11}}^\infty \frac1n\), not \(\sum_{n=10}^\infty \frac1n\). Plug in \(n=1\) to check: the first term is \(\frac1{11}\).
Geometric series
A geometric series has a constant ratio \(r\) between consecutive terms:
\[ \sum_{n=0}^\infty a r^n = a + ar + ar^2 + \cdots = \begin{cases} \dfrac{a}{1-r}, & |r| < 1 \\[8pt] \text{diverges}, & |r| \ge 1 \end{cases} \]
The most reliable way to remember it, whatever the index:
\[ \text{sum} = \frac{\text{first term}}{1 - \text{ratio}} \qquad (\text{when } |r| < 1) \]
Example. \(\displaystyle 3 + \frac34 + \frac3{16} + \frac3{64} + \cdots = \sum_{n=0}^\infty 3\left(\frac14\right)^n\), so \(r = \frac14 < 1\) and the sum is \(\dfrac{3}{1 - \frac14} = 4\).
Example. \(\displaystyle 2 + \frac{2e}{\pi} + \frac{2e^2}{\pi^2} + \cdots = \sum_{n=0}^\infty 2\left(\frac{e}{\pi}\right)^n\). Since \(e < \pi\), \(r = \frac{e}{\pi} \approx 0.865 < 1\), so it converges to
\[ \frac{2}{1 - e/\pi} = \frac{2\pi}{\pi - e} \]
Flip it, \(\sum 2\left(\frac{\pi}{e}\right)^n\), and \(r = \frac{\pi}{e} > 1\), so it diverges.
Geometric series from a recursive description
Suppose \(a_1 = 3\) and \(\dfrac{a_{n+1}}{a_n} = \dfrac12\). Constant ratio means geometric: \(a_n = 3\left(\frac12\right)^{n-1}\).
WeBWorK often wants the form \(\sum_{n=1}^\infty a r^n\) (index starting at \(1\), exponent exactly \(n\)). Then the first term is \(ar\), so \(a = \dfrac{a_1}{r} = 6\):
\[ \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty 6\left(\frac12\right)^n = \frac{\text{first term}}{1-r} = \frac{3}{1 - \frac12} = 6 \]
If instead \(a_1 = 2\) and \(\dfrac{a_{n+1}}{a_n} = -4\), then \(\sum a_n = \sum_{n=1}^\infty \left(-\tfrac12\right)(-4)^n\), and \(|r| = 4 \ge 1\), so it diverges.
\(\frac{a}{1-r}\) assumes the series starts at \(ar^0 = a\). If the series is \(\sum_{n=1}^\infty ar^n\), the first term is \(ar\) and the sum is \(\frac{ar}{1-r}\). Using “first term over \(1 - r\)” avoids this trap.
Practice: \(\displaystyle 5 - \frac53 + \frac59 - \frac5{27} + \cdots\); \(\;\displaystyle\sum_{n=1}^\infty \frac{2^{n+1}}{3^n}\); \(\;a_1 = 10,\ \frac{a_{n+1}}{a_n} = \frac{2}{5}\).
Functions as geometric series (a preview of power series)
Read the geometric sum formula backwards, with \(a = 1\) and \(r = y\):
\[ \frac{1}{1-y} = \sum_{n=0}^\infty y^n, \qquad |y| < 1 \]
Any function you can massage into \(\dfrac{(\text{stuff})}{1 - (\text{something})}\) becomes a series.
Example. \(\dfrac{5x}{1-x^2} = 5x \cdot \dfrac{1}{1 - x^2} = 5x\sum_{n=0}^\infty (x^2)^n = \displaystyle\sum_{n=0}^\infty 5x^{2n+1}\), for \(|x| < 1\).
Example. \(\dfrac{3x}{1+x}\). Rewrite the plus as minus a negative: \(1 + x = 1 - (-x)\).
\[ \frac{3x}{1+x} = 3x\sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty 3(-1)^n x^{n+1} = \sum_{n=1}^\infty 3(-1)^{n-1}x^n \]
Example. \(\dfrac{2}{1+\cos^2 x} = \displaystyle\sum_{n=0}^\infty 2(-1)^n\cos^{2n}(x)\), valid whenever \(\cos^2 x < 1\).
Practice: \(\dfrac{4}{1-x^3}\), \(\;\dfrac{2\sqrt{x}}{1 - x^{5/2}}\), \(\;\dfrac{x^2}{1 + 2x}\) (what’s the interval?)
Part 2: Convergence tests (5.3)
For most series we can’t find a formula for \(S_N\), so we can’t compute the sum. The tests below answer the easier question: does it converge at all?
The Divergence Test (\(n\)th-term test)
If \(\displaystyle\lim_{n\to\infty} a_n \ne 0\) (or the limit does not exist), then \(\displaystyle\sum a_n\) diverges.
Why: if the series converged, \(a_n = S_n - S_{n-1} \to S - S = 0\).
| Series | \(\lim a_n\) | Conclusion |
|---|---|---|
| \(\displaystyle\sum \frac{3n^2}{1+4n^2}\) | \(\frac34\) | diverges |
| \(\displaystyle\sum \frac{-5n}{\sqrt{4n^2+1}}\) | \(-\frac52\) | diverges |
| \(\displaystyle\sum \arctan(n)\) | \(\frac{\pi}{2}\) | diverges |
| \(\displaystyle\sum e^{1/n}\) | \(e^0 = 1\) | diverges |
| \(\displaystyle\sum (-1)^n\) | DNE | diverges |
| \(\displaystyle\sum \frac1n\) | \(0\) | test does not apply |
| \(\displaystyle\sum \frac1{n^2}\) | \(0\) | test does not apply |
\(\lim a_n = 0\) tells you nothing: \(\sum \frac1n\) diverges and \(\sum \frac1{n^2}\) converges, and both have terms going to \(0\). The Divergence Test can only ever conclude diverges. Never write “converges by the Divergence Test.”
For limits like \(\frac{-5n}{\sqrt{4n^2+1}}\), divide top and bottom by \(n = \sqrt{n^2}\): \(\dfrac{-5}{\sqrt{4 + 1/n^2}} \to \dfrac{-5}{2}\).
Practice: \(\displaystyle\sum \frac{n^3}{2-5n^3}\), \(\;\displaystyle\sum \cos\!\left(\frac1n\right)\), \(\;\displaystyle\sum \frac{n}{e^n}\)
\(p\)-series
\[ \sum_{n=1}^\infty \frac{1}{n^p} \quad \begin{cases} \text{converges}, & p > 1 \\ \text{diverges}, & p \le 1 \end{cases} \]
\(p = 1\) is the harmonic series, the boundary case, and it diverges.
The work is usually algebra to find \(p\): get the term into the form \(1/n^{p}\).
- \(\displaystyle\sum \frac{1}{n^{0.9}}\): \(p = 0.9 \le 1\), diverges.
- \(\displaystyle\sum \frac{1}{\sqrt[3]{n^4}} = \sum \frac{1}{n^{4/3}}\): \(p = \frac43 > 1\), converges.
- \(\displaystyle\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}\): \(p = \frac12\), diverges.
- \(\displaystyle\sum \frac{n^{\sqrt2}}{n^{3}} = \sum \frac{1}{n^{3-\sqrt2}}\): \(p = 3 - \sqrt2 \approx 1.59 > 1\), converges.
A disguised \(p\)-series. \(\displaystyle\sum_{n=1}^\infty 3^{-\ln n}\). Rewrite with base \(e\):
\[ 3^{-\ln n} = e^{-\ln n \cdot \ln 3} = \left(e^{\ln n}\right)^{-\ln 3} = n^{-\ln 3} = \frac{1}{n^{\ln 3}} \]
So \(p = \ln 3 \approx 1.099 > 1\): converges. In general \(b^{-\ln n} = \dfrac{1}{n^{\ln b}}\), so the question becomes whether \(\ln b > 1\), i.e. whether \(b > e\).
Practice: \(\displaystyle\sum \frac{1}{\sqrt[4]{n^3}}\), \(\;\displaystyle\sum \frac{n^{\pi}}{n^{5}}\), \(\;\displaystyle\sum 2.5^{-\ln n}\), \(\;\displaystyle\sum 10^{-\ln n}\)
The Integral Test
Suppose \(a_n = f(n)\), where \(f\) is positive, continuous, and decreasing on \([N, \infty)\). Then
\[ \sum_{n=N}^\infty a_n \quad\text{and}\quad \int_N^\infty f(x)\,dx \]
either both converge or both diverge.
The picture: rectangles of width \(1\) and height \(a_n\) sit under (or over) the curve \(y = f(x)\), so the series and the integral are trapped together.
If \(\int_1^\infty f(x)\,dx = \frac{\pi}{8}\), that tells you the series converges. It does not tell you the series equals \(\frac{\pi}{8}\).
Example (diverges). \(\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt[3]{n+2}}\)
\[ \int_1^\infty (x+2)^{-1/3}\,dx = \lim_{t\to\infty} \frac32 (x+2)^{2/3}\Big|_1^t = \infty \]
Diverges. (You could also argue it behaves like a \(p\)-series with \(p = \frac13\).)
Example (\(u\)-sub to arctan). \(\displaystyle\sum_{n=0}^\infty \frac{n}{4+n^4}\)
Let \(u = x^2\), \(du = 2x\,dx\):
\[ \int_0^\infty \frac{x}{4+x^4}\,dx = \frac12\int_0^\infty \frac{du}{4+u^2} = \frac12\cdot\frac12\arctan\!\left(\frac u2\right)\Big|_0^\infty = \frac14\cdot\frac\pi2 = \frac\pi8 \]
Finite, so the series converges. (Strictly, \(f(x) = \frac{x}{4+x^4}\) is decreasing only once \(x > (4/3)^{1/4} \approx 1.07\). “Eventually decreasing” is enough, because the first few terms don’t affect convergence.)
Example (exponentials). \(\displaystyle\sum_{n=0}^\infty \frac{e^{2n}}{1+e^{4n}}\)
Let \(u = e^{2x}\), \(du = 2e^{2x}\,dx\); when \(x = 0\), \(u = 1\):
\[ \int_0^\infty \frac{e^{2x}}{1 + e^{4x}}\,dx = \frac12\int_1^\infty \frac{du}{1+u^2} = \frac12\left(\frac\pi2 - \frac\pi4\right) = \frac\pi8 \]
Converges.
Example (logs). \(\displaystyle\sum_{n=2}^\infty \frac{1}{n\ln n}\) vs. \(\displaystyle\sum_{n=2}^\infty \frac{1}{n(\ln n)^3}\). Let \(u = \ln x\), \(du = \frac{dx}{x}\):
\[ \int_2^\infty \frac{dx}{x\ln x} = \int_{\ln 2}^\infty \frac{du}{u} = \ln u\Big|_{\ln 2}^\infty = \infty \quad\Rightarrow\quad \text{diverges} \]
\[ \int_2^\infty \frac{dx}{x(\ln x)^3} = \int_{\ln 2}^\infty u^{-3}\,du = -\frac{1}{2u^2}\Big|_{\ln2}^\infty = \frac{1}{2(\ln 2)^2} \quad\Rightarrow\quad \text{converges} \]
After the substitution these are just \(p\)-integrals in \(u\), so the same \(p > 1\) rule decides them.
Practice: \(\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt[4]{n+5}}\), \(\;\displaystyle\sum_{n=0}^\infty \frac{n}{1+n^4}\), \(\;\displaystyle\sum_{n=2}^\infty \frac{1}{n\sqrt{\ln n}}\), \(\;\displaystyle\sum_{n=1}^\infty n e^{-n^2}\)
Estimating the remainder
If \(\sum a_n\) converges by the Integral Test, adding up the first \(N\) terms leaves an error, the remainder
\[ R_N = \sum_{n=1}^\infty a_n - \sum_{n=1}^N a_n = a_{N+1} + a_{N+2} + \cdots \]
and the same rectangle picture gives
\[ R_N \le \int_N^\infty f(x)\,dx \]
Example. \(\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}\)
\[ R_N \le \int_N^\infty x^{-4}\,dx = \frac{1}{3N^3}, \qquad R_{20} \le \frac{1}{3\cdot 20^3} = \frac{1}{24000} \approx 4.2\times 10^{-5} \]
Example. \(\displaystyle\sum_{n=1}^\infty \frac{1}{1+n^2}\)
\[ R_N \le \int_N^\infty \frac{dx}{1+x^2} = \frac{\pi}{2} - \arctan(N), \qquad R_{100} \le \frac\pi2 - \arctan(100) \approx 0.0099997 \]
Running it backwards: how many terms do I need? To approximate \(\sum \frac1{n^4}\) with error less than \(10^{-6}\), solve
\[ \frac{1}{3N^3} < 10^{-6} \iff N^3 > 333{,}333.\overline{3} \iff N \ge 70 \]
(since \(69^3 = 328{,}509\) is too small and \(70^3 = 343{,}000\) is enough).
Practice: bound \(R_N\) and \(R_{30}\) for \(\displaystyle\sum \frac{1}{n^5}\); bound \(R_{500}\) for \(\displaystyle\sum \frac{2}{4+n^2}\).
Which tool do I reach for?
- Do the terms go to \(0\)? If not, stop: it diverges by the Divergence Test.
- Is it geometric (constant ratio)? Check \(|r| < 1\); if so, the sum is \(\frac{\text{first term}}{1-r}\).
- Is it a \(p\)-series, maybe after simplifying exponents or rewriting \(b^{-\ln n}\)? Check \(p > 1\).
- Does it telescope (partial fractions, \(f(n) - f(n+1)\), log of a quotient)? Find \(S_k\) and take the limit.
- Is the term something you can integrate (often a \(u\)-sub to \(\arctan\), \(\ln\), or a power)? Use the Integral Test.
Only geometric and telescoping series hand you an actual value. The Divergence Test, \(p\)-series, and Integral Test only tell you converges or diverges.