MA1101 / MATHEMATICS FOR ELECTRONICS I

Part IV / Ch 10

MA1101ContentsClose ×CH 10

Part IV · Series, Exponentials, and Integral Calculus

10

Chapter 10 — Infinite Series and Convergence Tests

Why This Matters

Why the ratio test works — and when it does not. The ratio test checks whether the tail of a series eventually behaves like a convergent geometric series by measuring the limiting ratio between consecutive terms.

The ratio test as geometric comparison

Starting point. Let \(\sum a_n\) be a series with non-zero terms, and let \(L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\).

  1. \[|a_{n+1}| \approx L |a_n| \quad \text{for large } n\]
    For large \(n\), consecutive terms scale by a factor close to \(L\). The tail approximates a geometric progression with ratio \(L\).reversible
  2. \[|a_{N+k}| \approx L^k |a_N|\]
    After \(k\) additional steps past index \(N\), the term magnitude scales as \(L^k |a_N|\).reversible
  3. \[L < 1 \implies \text{converges}, \quad L > 1 \implies \text{diverges}\]
    When \(L < 1\), the geometric series \(\sum L^k\) converges, bounding the tail. When \(L > 1\), terms grow without bound.reversible

Conclusion. The ratio test compares the series tail to a geometric series. When \(L < 1\), terms decay exponentially fast; when \(L > 1\), terms do not approach zero.

An infinite expression is meaningful only when its partial results settle to a definite value. Series make it possible to define and approximate functions that finite polynomials cannot express exactly, including the exponential function that governs capacitor transients.

10.1 Geometric series

The finite geometric sum is

\[S_N=1+r+r^2+\cdots+r^N.\]

Multiply by \(r\) and subtract:

\[\begin{aligned} S_N-rS_N&=(1+r+\cdots+r^N)-(r+r^2+\cdots+r^{N+1})\\ &=1-r^{N+1}. \end{aligned}\]

Therefore, for \(r\ne1\),

\[S_N=\frac{1-r^{N+1}}{1-r}.\]

If \(|r|<1\), then \(r^{N+1}\to0\) and

\[\sum_{n=0}^{\infty}r^n=\frac1{1-r}.\]

If \(|r|\ge1\), the terms do not tend to zero in the required way, so the series diverges. The condition is part of the formula.

Engineering Connection — repeated attenuation

If each stage passes a fraction \(r\) of a quantity and \(|r|<1\), accumulated contributions can have the form \(A(1+r+r^2+\cdots)=A/(1-r)\). The model is valid only when the repeated effect really is linear and each stage has the same ratio.

Derivation. Multiply a finite geometric sum by its ratio and subtract. Every middle term cancels, leaving only the first term and one tail term. Infinity enters only after this finite identity is established.

\[S_n=a\frac{1-r^n}{1-r}\quad(r\ne1)\]

10.2 Positive-term comparison and \(p\)-series

For nonnegative terms, comparison turns a difficult series into a known one. If

\[0\le a_n\le b_n\]

eventually and \(\sum b_n\) converges, then \(\sum a_n\) converges. If \(a_n\ge b_n\ge0\) eventually and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.

The \(p\)-series gives a central benchmark:

\[\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges},&p>1,\\ \text{diverges},&p\le1. \end{cases}\]

The harmonic series is the \(p=1\) case. Its terms approach zero but its partial sums do not settle, demonstrating again that term limit zero is not sufficient.

Why \(p = 1\) is the boundary for convergence. The \(p\)-series \(\sum 1/n^p\) is the central reference benchmark for comparison tests. The threshold \(p = 1\) reflects the balance point where the continuous improper integral transitions from finite to infinite area.

Integral test for p-series

Starting point. Let \(f(x) = \frac{1}{x^p}\) be positive and decreasing on \([1, \infty)\).

  1. \[\int_1^{n+1} f(x)\,dx \le \sum_{k=1}^n f(k) \le f(1) + \int_1^n f(x)\,dx\]
    The sum of rectangle areas is sandwiched between two shifted continuous integrals.reversible
  2. \[\int_1^\infty \frac{dx}{x^p} < \infty \iff p > 1\]
    The improper integral converges if and only if \(p > 1\), since \(\lim_{b\to\infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b\) is finite only when \(1-p < 0\).reversible
  3. \[\sum_{n=1}^\infty \frac{1}{n^p} < \infty \iff p > 1\]
    Since the partial sums are bounded by the convergent integral, the \(p\)-series converges exactly when \(p > 1\).reversible

Conclusion. The \(p\)-series \(\sum_{n=1}^\infty \frac{1}{n^p}\) converges if and only if \(p > 1\). The harmonic series (\(p = 1\)) sits right on the boundary and diverges.

  1. Identify the series as a \(p\)-series \(\sum \frac{1}{n^p}\) or compare it to one using the Limit Comparison Test.
  2. If \(p > 1\): the series converges (useful for establishing upper bounds).
  3. If \(p \le 1\): the series diverges (useful for establishing lower bounds to prove divergence).
  4. Remember: \(p = 1.0001\) converges, while \(p = 0.9999\) diverges. The boundary at \(p = 1\) is sharp.

10.3 Ratio and root tests

For a series \(\sum a_n\), the ratio test examines

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

If \(L<1\), the series converges absolutely; if \(L>1\) or \(L=\infty\), it diverges; if \(L=1\), the test says nothing. The root test has the corresponding quantity

\[L=\lim_{n\to\infty}|a_n|^{1/n}.\]

It has the same conclusions. Factorials often make ratios simple; powers raised to the \(n\)th power often make roots simple.

Example — Factorials favour the ratio test

Study \(\sum_{n=1}^{\infty}\frac{3^n}{n!}\). Then

\[\left|\frac{a_{n+1}}{a_n}\right| =\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac3{n+1}\to0.\]

Because \(0<1\), the series converges absolutely.

The ratio test is an eventual comparison. Once consecutive absolute terms are consistently reduced by a factor below one, the remaining tail is dominated by a convergent geometric tail. When the limiting ratio is one, that comparison becomes neutral; it says nothing about the series.

10.4 Power series and radius of convergence

A power series centred at \(0\) has form

\[\sum_{n=0}^{\infty}a_nx^n.\]

There is a radius \(R\in[0,\infty]\) such that the series converges absolutely for \(|x|<R\) and diverges for \(|x|>R\). At \(|x|=R\), each endpoint must be tested separately. When the relevant root limit exists,

\[\frac1R=\lim_{n\to\infty}|a_n|^{1/n}.\]

For a series centred at \(a\), replace \(x\) by \(x-a\) throughout.

Why endpoints require their own work. The ratio test describes an open interval where the terms shrink decisively and an exterior where they do not. At the boundary its diagnostic often becomes exactly one, which gives no conclusion. Each endpoint is a new series and must be tested on its own terms.

10.5 Functions defined by power series

Within its radius of convergence, a power series is continuous and may be differentiated term by term. The derivative has the same radius of convergence:

\[\frac d{dx}\sum_{n=0}^{\infty}a_nx^n =\sum_{n=1}^{\infty}na_nx^{n-1}.\]

This theorem is powerful precisely because it is restricted to the open interval of convergence. It is not an invitation to differentiate a divergent expression.

Why power series always converge on symmetric intervals. The behaviour of \(\sum a_n (x-c)^n\) depends on \(|x-c|\) — the absolute geometric distance from the centre \(c\) — rather than on the sign of \(x-c\).

Computing R from the ratio test

From
  • Apply the ratio test: \(\lim_{n\to\infty} \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x-c| \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = |x-c| L\).
  • Convergence requires \(|x-c| L < 1\).
  1. Solving for distance: \(|x-c| < \frac{1}{L}\).
  2. Define the radius of convergence \(R = \frac{1}{L}\) (with \(R = \infty\) if \(L = 0\), and \(R = 0\) if \(L = \infty\)).

Result. The series converges absolutely for \(|x-c| < R\) and diverges for \(|x-c| > R\), forming an interval centred at \(c\).

Limitation. The endpoints \(x = c - R\) and \(x = c + R\) must be tested separately using numerical series convergence tests.

Common Mistakes

  • Treating a sequence and its series as the same object.
  • Concluding a series converges solely because its terms tend to zero.
  • Reporting “ratio test equals one” as convergence or divergence.
  • Forgetting that the radius test does not decide the endpoints.

Fundamental distinction — sequence versus series. A sequence \(a_1, a_2, a_3, \ldots\) is an ordered list of numbers; each term exists independently. A series \(\sum a_n\) adds those numbers into a single accumulated value. Confusing the two is like confusing a shopping list with the total bill. The sequence \(1/n\) approaches \(0\), but the harmonic series \(\sum 1/n\) diverges — terms shrink, yet there are infinitely many of them.

Partial sums define the series

Starting point. Let \(S_N = \sum_{n=1}^N a_n\) be the \(N\)-th partial sum.

  1. \[S_1 = a_1\]
    The first partial sum is just the first term.reversible
  2. \[S_2 = a_1 + a_2\]
    The second partial sum adds the second term.reversible
  3. \[S_N = S_{N-1} + a_N\]
    Each new partial sum adds one more term to the previous accumulation.reversible
  4. \[\sum_{n=1}^\infty a_n = \lim_{N\to\infty} S_N\]
    The infinite series is defined as the limit of the partial-sum sequence \((S_N)\).reversible

Conclusion. Convergence of a series means the sequence of partial sums \((S_N)\) settles to a finite number \(S\). It is not merely about individual terms \(a_n\) shrinking.

  1. First check the \(n\)-th term: if \(\lim_{n\to\infty} a_n \neq 0\), the series diverges immediately (Divergence Test).
  2. If \(\lim_{n\to\infty} a_n = 0\), the test is inconclusive — you must use a convergence test (Comparison, Integral, Ratio, Root, or Alternating Series).
  3. Never conclude a series converges just because \(a_n \to 0\).

10.6 Partial sums, test selection, and endpoint discipline

A series converges only when its partial sums settle. The term test is a fast necessary-condition check: if \(a_n\) does not tend to zero, \(\sum a_n\) diverges immediately. Passing this check proves nothing further. Choose comparison for positive rational-looking terms, ratio for factorials or exponentials, and root for an expression naturally raised to the \(n\)th power.

Example — finite geometric sum before the infinite limit

For \(1+1/2+\cdots+(1/2)^4\), the finite formula gives

\[S_4=\frac{1-(1/2)^5}{1-1/2}=\frac{31}{16}.\]

The infinite sum is \(2\), not because five terms equal it, but because the remainder \((1/2)^{N+1}/(1-1/2)\) tends to zero as \(N\) grows.

Example — comparison with a \(p\)-series

For \(n\ge1\),

\[0<\frac1{n^2+n}\le\frac1{n^2}.\]

Since \(\sum1/n^2\) converges, \(\sum1/(n^2+n)\) converges. The inequality direction matters: a smaller nonnegative series inherits convergence from a larger known convergent series, not the other way around.

Example — radius is not an interval

For \(\sum_{n=1}^{\infty}(x-1)^n/n\), the ratio test gives \(|x-1|<1\). At \(x=2\), the series is harmonic and diverges; at \(x=0\), it is alternating harmonic and converges. The interval is \([0,2)\), not simply “radius one.”

A systematic approach to choosing convergence tests. With several convergence tests available, the essential skill is matching the test to the algebraic structure of the summand.

  1. <strong>Divergence Test first:</strong> if \(\lim_{n\to\infty} a_n \neq 0\), the series diverges immediately. Done.
  2. <strong>Factorials (\(n!\)) or products:</strong> use the Ratio Test, since factorials simplify cleanly in ratios \(\frac{(n+1)!}{n!} = n+1\).
  3. <strong>\(n\)-th powers (\(a_n = (b_n)^n\)):</strong> use the Root Test to eliminate the outer exponent.
  4. <strong>Rational or algebraic functions of \(n\):</strong> use the Limit Comparison Test with a \(p\)-series \(\sum 1/n^p\).
  5. <strong>Alternating signs with \(|a_n|\) decreasing to \(0\):</strong> use the Alternating Series Test.
  6. <strong>Continuous, positive, decreasing function \(f(n) = a_n\):</strong> use the Integral Test.
  7. <strong>Power series:</strong> use the Ratio or Root Test to find the radius \(R\), then test each endpoint \(x = c \pm R\) separately.

Endpoint testing is independent. The Ratio or Root Test yields the open interval \((c-R, c+R)\). At \(x = c+R\) and \(x = c-R\), substitute the numerical value and test each resulting series on its own merits. One endpoint may converge while the other diverges.

10.7 Choosing a convergence test and representing functions

Tests are tools with limited conclusions. A ratio or root limit equal to \(1\) is not a near-convergence result; it is no result at all from that test. For a power series, first find the open radius interval, then test each endpoint in the original series. Endpoint behaviour can differ because signs or known benchmark series change there.

Example — factorials signal the ratio test

For

\[\sum_{n=1}^{\infty}\frac{n!}{4^n},\]

the term ratio is

\[\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)!/4^{n+1}}{n!/4^n} =\frac{n+1}{4}\to\infty.\]

The terms eventually grow rather than approach zero, so the series diverges. The ratio test was chosen because factorial cancellation reveals the behaviour in one step.

Example — an \(n\)th power signals the root test

For

\[\sum_{n=1}^{\infty}\left(\frac{2n+1}{3n+4}\right)^n,\]

the root test gives

\[\lim_{n\to\infty}\left|a_n\right|^{1/n} =\lim_{n\to\infty}\frac{2n+1}{3n+4}=\frac23<1.\]

Therefore the series converges absolutely. A ratio test could also work but would require more algebra; the root test matches the expression's structure.

Example — power-series endpoints are separate problems

For

\[\sum_{n=1}^{\infty}\frac{x^n}{n\,2^n},\]

the ratio test gives \(|x|/2<1\), so \(|x|<2\). At \(x=2\) the series is harmonic and diverges. At \(x=-2\) it is alternating harmonic and converges. Thus the convergence interval is \([-2,2)\).

Example — a power series represents a local function

Within \(|x|<1\),

\[\frac1{1-x}=1+x+x^2+x^3+\cdots.\]

At \(x=0.1\), the first four terms give \(1.111\), while the exact value is \(1/0.9=1.111\overline1\). The omitted tail is geometric and has size \(0.1^4/(1-0.1)\), so its smallness is quantified. At \(x=1.1\), the rational function exists but this series diverges: a function and one particular power series representation need not share the same domain.

Why the geometric series formula holds. The sum formula is not an approximation — it comes from exact algebraic cancellation (a telescoping identity) followed by a limit.

Algebraic derivation of the geometric sum

Starting point. Let \(S_n = \sum_{k=0}^{n-1} a r^k = a + ar + ar^2 + \cdots + ar^{n-1}\).

  1. \[S_n = a + ar + ar^2 + \cdots + ar^{n-1}\]
    Write out the finite sum explicitly.reversible
  2. \[r S_n = ar + ar^2 + \cdots + ar^n\]
    Multiply every term by the common ratio \(r\).reversible
  3. \[S_n - r S_n = a - ar^n\]
    Subtract the second line from the first: intermediate terms \(ar, ar^2, \dots, ar^{n-1}\) cancel in pairs.reversible
  4. \[(1-r) S_n = a(1 - r^n)\]
    Factor out \(S_n\) on the left and \(a\) on the right.reversible
  5. \[S_n = \frac{a(1-r^n)}{1-r} \quad (r \neq 1)\]
    Divide by \(1-r\). For \(|r| < 1\), as \(n \to \infty\), \(r^n \to 0\), giving \(S_\infty = \frac{a}{1-r}\).reversible

Conclusion. The finite geometric sum is \(S_n = \frac{a(1-r^n)}{1-r}\). When \(|r| < 1\), the infinite sum converges to \(S = \frac{a}{1-r}\).

The r = 1 edge case

From
  • The derivation divides by \(1-r\), which is zero when \(r = 1\).
  • When \(r = 1\), every term in the sum equals \(a\).
  1. If \(r = 1\): \(S_n = a + a + \cdots + a = n a\).
  2. As \(n \to \infty\), \(n a \to \infty\) (for \(a \neq 0\)).

Result. The geometric series with \(r = 1\) diverges to infinity. The edge case requires separate arithmetic because division by \(1-r\) is undefined.

Limitation. When \(r = 1\), multiplying by \(r\) changes nothing, so subtraction yields \(0 = 0\) rather than a formula.

Exercises

  1. Find \(\sum_{n=0}^{\infty}(1/3)^n\).
  2. Decide whether \(\sum_{n=1}^{\infty}1/(n^2+1)\) converges by comparison.
  3. Use the ratio test for \(\sum n!/5^n\).
  4. Find the radius and test endpoints for \(\sum_{n=1}^{\infty}x^n/n\).
  5. Explain why \(\sum x^n=1/(1-x)\) cannot be used at \(x=2\).
  6. Find the sum of the first six terms of \(1+1/3+1/9+\cdots\) and its infinite sum.
  7. Decide convergence of \(\sum1/(n^2+n)\) by comparison.
  8. Find the convergence interval of \(\sum(x-1)^n/n\).
  9. Choose and apply an appropriate test to \(\sum n!/4^n\).
  10. Choose and apply an appropriate test to \(\sum((2n+1)/(3n+4))^n\).
  11. Find the interval of convergence of \(\sum x^n/(n\,2^n)\), testing both endpoints.