Unit 3

Sequences & Infinite Series

Can you add infinitely many numbers and get a finite answer? Sometimes. This unit is a toolkit of tests for deciding which — and for an electrical engineer that question is not academic. Whether a series converges is whether a feedback loop is stable and whether a digital filter blows up.

1 · Sequences

A sequence is an ordered list \(a_1, a_2, a_3, \ldots\) — formally, a function whose domain is the positive integers. It converges to \(L\) if the terms get and then stay arbitrarily close to \(L\).

You already know how to do this

To find \(\lim a_n\), treat \(n\) as a continuous variable and use every limit tool from Calc I — including L'Hôpital's rule, which needs a continuous variable to be legal in the first place. If \(f(x) \to L\) and \(a_n = f(n)\), then \(a_n \to L\).

Limits worth knowing on sight

Growth ordering, slowest to fastest: \(\ln n \;<\; n^p \;<\; a^n \;<\; n! \;<\; n^n\). Memorise this chain — it settles most limit comparisons instantly.
Sequence	Limit	Note
\(r^n\)	\(0\) if \(\|r\|<1\); \(1\) if \(r=1\); diverges otherwise	The most important one in EE
\(n^{1/n}\)	\(1\)	Take logs to see it
\(\left(1 + \dfrac{x}{n}\right)^n\)	\(e^{x}\)	The definition of \(e\)
\(\dfrac{x^n}{n!}\)	\(0\), for every \(x\)	Factorial beats exponential, always
\(\dfrac{\ln n}{n^p}\)	\(0\), for every \(p>0\)	Any positive power beats a logarithm

Monotone Convergence Theorem: a sequence that is monotone (always increasing or always decreasing) and bounded must converge. You do not need to know the limit to prove it exists — useful for recursively defined sequences where no closed form is available.

2 · Series and partial sums

A series is what you get by adding a sequence. Its meaning is defined through the partial sums:

\[ s_N = \sum_{n=1}^{N} a_n \qquad\qquad \sum_{n=1}^{\infty} a_n = \lim_{N\to\infty} s_N \] The infinite sum is the limit of the finite sums. If that limit exists and is finite, the series converges; otherwise it diverges. There is no other definition — every test in this unit is ultimately a shortcut for evaluating this.

Sequence vs series — keep them apart

The sequence \(1/n\) converges, to \(0\). The series \(\sum 1/n\) diverges, to infinity. These are different objects and the words are nearly identical, which is exactly why exams test the distinction. Terms shrinking to zero is necessary for a series to converge but nowhere near sufficient.

3 · Geometric series

The one series you can always sum exactly, and the one an engineer uses constantly:

\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \cdots = \frac{a}{1-r}, \qquad |r| < 1 \] \(a\) is the first term actually present and \(r\) the common ratio. If the sum starts at \(n=1\) with \(ar^n\), the first term is \(ar\), so the sum is \(ar/(1-r)\). Getting \(a\) wrong is the standard error.

Where the formula comes from, in one line:

\[ s_N = \frac{a\big(1 - r^{N}\big)}{1-r}, \qquad r^{N} \to 0 \iff |r| < 1 \]

Everything about geometric series is that one condition.

\(1 + \tfrac12 + \tfrac14 + \cdots\) with \(a=1\), \(r=\tfrac12\), summing to \(1/(1-\tfrac12) = 2\). Bars are the individual terms; dots are the running totals. The terms must shrink fast enough that the totals level off.

EE: \(|r| < 1\) is a stability criterion

Feedback. A loop with forward gain \(A\) and feedback fraction \(\beta\) sends a signal around and around, contributing \(A\), then \(-A^2\beta\), then \(+A^3\beta^2\), … — a geometric series with \(r = -A\beta\):

\[ A\Big(1 - A\beta + (A\beta)^2 - \cdots\Big) = \frac{A}{1 + A\beta} \]

That is the closed-loop gain formula from your circuits course, and the derivation is valid precisely when \(|A\beta| < 1\).

Digital filters. A first-order IIR filter \(y[n] = x[n] + a\,y[n-1]\) has impulse response \(a^n\). Its total response to a bounded input is \(\sum a^n\), which converges only for \(|a| < 1\). "Poles inside the unit circle" is literally the geometric series convergence condition — you are meeting the stability criterion here, two years early.

Telescoping series

If \(a_n\) can be written as a difference \(b_n - b_{n+1}\) — partial fractions from Unit 1 is how you find this — then almost everything cancels:

\[ \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = \sum_{n=1}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 \] \(s_N = 1 - \dfrac{1}{N+1} \to 1\). Always write out the partial sum to see what survives — usually the first term and a vanishing tail.

4 · The \(n\)th-term test

\[ \text{If } \lim_{n\to\infty} a_n \neq 0, \text{ then } \sum a_n \textbf{ diverges.} \] This is the only thing this test can conclude. If the limit is \(0\), the test is inconclusive — it tells you nothing at all.

Run it first on every series; it is free. It is the fastest way to dispose of things like \(\sum \frac{n}{2n+1}\), whose terms head to \(\tfrac12\).

The single most common error in this unit

Concluding that \(\sum a_n\) converges because \(a_n \to 0\). The harmonic series \(\sum 1/n\) is the standing counterexample: its terms go to zero and it still diverges to infinity. Writing "converges by the \(n\)th-term test" earns zero marks, because there is no such conclusion.

5 · Integral test and p-series

If \(f\) is continuous, positive and decreasing on \([1,\infty)\) with \(f(n) = a_n\), then \(\sum a_n\) and \(\int_1^{\infty} f(x)\,dx\) both converge or both diverge.

All three conditions matter. The picture is that the series is a Riemann-sum sandwich around the integral: rectangles of width 1 either all sit under the curve or all sit over it, so their total and the area share a fate.

\[ \sum_{n=1}^{\infty}\frac{1}{n^{p}} \text{ converges} \iff p > 1 \] Directly from the p-integral in Unit 1. \(p=1\) is the harmonic series — divergent, and the borderline you must know.

Remainder estimate

The integral test also bounds the error when you truncate:

\[ \int_{N+1}^{\infty} f(x)\,dx \;\le\; R_N \;\le\; \int_{N}^{\infty} f(x)\,dx \]

That answers "how many terms do I need for three-digit accuracy?" — a question that matters the moment you compute a series numerically rather than symbolically.

6 · Comparison tests

For series with positive terms only.

Direct comparison

If \(0 \le a_n \le b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
If \(0 \le b_n \le a_n\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

Smaller than convergent ⇒ convergent. Bigger than divergent ⇒ divergent. The other two combinations conclude nothing — being smaller than something infinite is no information.

Limit comparison — usually the better tool

\[ \text{If } \lim_{n\to\infty}\frac{a_n}{b_n} = c \text{ with } 0 < c < \infty, \text{ then } \sum a_n \text{ and } \sum b_n \text{ behave identically.} \] No inequality to establish — just a limit. This is why limit comparison is easier in practice than direct comparison, where finding a valid bound can be fiddly.

How to pick \(b_n\)

Keep only the dominant power in numerator and denominator. For

\[ a_n = \frac{3n^2+5}{n^4 - 2n + 7} \]

the leading behaviour is \(3n^2/n^4 = 3/n^2\), so compare against \(b_n = 1/n^2\) — a convergent p-series with \(p=2\), so the original converges. This handles essentially every rational-function series you will see.

Even faster: subtract the numerator's degree from the denominator's. Here \(4 - 2 = 2\), so it behaves like a p-series with \(p=2\). Convergent whenever that difference exceeds 1.

7 · Alternating series

For \(\sum (-1)^n b_n\) with \(b_n > 0\), the series converges if both:

\(b_n\) is decreasing (eventually), and
\(\lim b_n = 0\).

\[ |R_N| = |S - s_N| \;\le\; b_{N+1} \] The error is smaller than the first term you left out. Simplest and sharpest error bound in the whole unit — and it is why alternating series are so pleasant to compute with.

8 · Absolute vs conditional convergence

Absolute convergence implies convergence. The converse fails — that is exactly what "conditional" means.
\(\sum a_n\)	\(\sum \|a_n\|\)	Called	Example
Converges	Converges	Absolutely convergent	\(\sum (-1)^n/n^2\)
Converges	Diverges	Conditionally convergent	\(\sum (-1)^n/n\)
Diverges	Diverges	Divergent	\(\sum (-1)^n\)

Why the distinction is not pedantry

A conditionally convergent series can be rearranged to sum to any number you like — Riemann's rearrangement theorem. Its value depends on the order of addition. Absolutely convergent series can be reordered freely with no change.

For an engineer that is a warning about numerics: summing a conditionally convergent series in a different order — or in floating point, where the machine effectively reorders through rounding — can give a different answer. Absolute convergence is what makes a computation trustworthy.

9 · Ratio and root tests

\[ L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \qquad\qquad L = \lim_{n\to\infty}\sqrt[n]{|a_n|} \]

For either test:

\(L < 1\) ⇒ converges absolutely
\(L > 1\) (or \(\infty\)) ⇒ diverges
\(L = 1\) ⇒ inconclusive — use another test

Both are really asking the same question: is this eventually geometric with ratio less than 1? Use the ratio test when you see factorials or \(n\) in an exponent; use the root test when the whole term is raised to the \(n\)th power.

\(L = 1\) happens exactly when you need it not to

The ratio test always gives \(L = 1\) on p-series, so it can never resolve \(\sum 1/n\) versus \(\sum 1/n^2\). Those need the integral test or a p-series comparison. Knowing when a test cannot help saves real time in an exam.

10 · Which test to use

Work down this list. Stop at the first line that applies.

Do the terms go to zero? If not ⇒ diverges by the \(n\)th-term test. Done.
Is it geometric or telescoping? ⇒ sum it exactly.
Is it a p-series (or obviously like one)? ⇒ \(p>1\) converges.
Factorials or \(c^n\)? ⇒ ratio test.
Whole term an \(n\)th power? ⇒ root test.
A rational function of \(n\), or algebraic-looking? ⇒ limit comparison against the dominant power.
Does it alternate? ⇒ check absolute convergence first; if that fails, alternating series test for conditional convergence.
Is \(a_n = f(n)\) with \(f\) positive, decreasing and easy to integrate? ⇒ integral test.

11 · Worked examples

1. Geometric series — read \(a\) and \(r\) carefully

Evaluate \(\displaystyle\sum_{n=2}^{\infty} 3\left(\tfrac25\right)^n\).

The trap is the starting index. The first term present is the \(n=2\) term:

\[ a = 3\left(\tfrac25\right)^2 = 3\cdot\tfrac{4}{25} = \tfrac{12}{25}, \qquad r = \tfrac25 \] \[ S = \frac{a}{1-r} = \frac{12/25}{3/5} = \frac{12}{25}\cdot\frac53 = \frac45 \]

Since \(|r| = 0.4 < 1\) it converges. Starting the sum at \(n=0\) instead would have given \(3/(1-0.4) = 5\) — a completely different answer.

2. Limit comparison

Determine whether \(\displaystyle\sum \frac{3n^2+5}{n^4-2n+7}\) converges.

Find the dominant behaviour: \(3n^2/n^4 = 3/n^2\). Compare with \(b_n = 1/n^2\).

\[ \lim_{n\to\infty}\frac{(3n^2+5)/(n^4-2n+7)}{1/n^2} = \lim_{n\to\infty}\frac{3n^4+5n^2}{n^4-2n+7} = 3 \]

Finite and nonzero, so the two series share a fate. \(\sum 1/n^2\) is a p-series with \(p = 2 > 1\), hence convergent. Therefore the original converges.

Why not direct comparison? The denominator \(n^4-2n+7\) is smaller than \(n^4\) for small \(n\), making the fraction bigger and breaking the inequality you would want. Limit comparison sidesteps that entirely — which is why it is the default.

3. Ratio test with a factorial

Determine whether \(\displaystyle\sum \frac{n!}{n^n}\) converges.

\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} \]

Use \((n+1)! = (n+1)\,n!\) and cancel:

\[ = \frac{(n+1)\,n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^{n} = \left(1+\frac1n\right)^{-n} \] \[ L = \lim_{n\to\infty}\left(1+\frac1n\right)^{-n} = \frac1e \approx 0.368 < 1 \]

Converges. Note where the standard limit \((1+1/n)^n \to e\) did the work — this is why that limit is on the memorise list.

4. Alternating series and truncation error

Show that \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}\) converges, and find how many terms guarantee an error below \(10^{-3}\).

Convergence. \(b_n = 1/n^3\) is positive, decreasing, and \(\to 0\). Alternating series test: converges. It is in fact absolutely convergent, since \(\sum 1/n^3\) is a p-series with \(p=3\).

Error. The remainder is bounded by the first omitted term:

\[ |R_N| \le b_{N+1} = \frac{1}{(N+1)^3} < 10^{-3} \implies (N+1)^3 > 1000 \implies N \ge 10 \]

Ten terms for three decimal places. Compare the alternating harmonic series, where \(b_{N+1} = 1/(N+1)\) would need a thousand terms for the same accuracy. How fast the terms shrink controls the cost of computing, not just whether it converges.

5. EE — stability of a feedback loop

An amplifier has open-loop gain \(A = 1000\) and feedback fraction \(\beta = 0.01\). Write the closed-loop gain as a geometric series, evaluate it, and state when the series representation is valid.

Loop gain: \(A\beta = 1000(0.01) = 10\).

Signal circulating the loop contributes \(A\big[1 - A\beta + (A\beta)^2 - \cdots\big]\), a geometric series with \(r = -A\beta = -10\).

\[ |r| = 10 > 1 \implies \text{the series } \textbf{diverges} \]

So is the amplifier unstable? No — and this is the subtle part worth getting right. The closed-loop gain formula

\[ A_{\text{CL}} = \frac{A}{1+A\beta} = \frac{1000}{11} \approx 90.9 \]

is still correct; it just cannot be derived by summing that series here. The situation mirrors \(1/(1-x)\): the function is defined for all \(x \neq 1\), but its series representation only converges for \(|x| < 1\). The formula extends beyond where the series does — a theme that returns as analytic continuation in Unit 4.

What genuinely signals instability is loop gain reaching \(-1\) — magnitude 1 with 180° of phase shift — which makes the denominator zero. That is the Barkhausen criterion, and it is how you deliberately build an oscillator rather than accidentally build one.

Checklist

I never conclude convergence from \(a_n \to 0\).
I identify \(a\) as the first term actually present, whatever the starting index.
I know \(\sum 1/n\) diverges and \(\sum 1/n^2\) converges, and why.
I reach for the ratio test when I see factorials, and limit comparison for rational functions.
I know the ratio test returns 1 and is useless on p-series.
I check absolute convergence before declaring a series conditionally convergent.
I can bound an alternating series' error with the first omitted term.
I can explain why \(|r| < 1\) is a stability condition in a feedback loop.