Which of the following statements about infinite series are correct?

If Σaₙ converges then aₙ → 0

The alternating harmonic series 1 − 1/2 + 1/3 − ... converges conditionally

The ratio test is inconclusive when the limit of |aₙ₊₁/aₙ| equals 1

If Σaₙ converges then aₙ → 0

The alternating harmonic series 1 − 1/2 + 1/3 − ... converges conditionally

The ratio test is inconclusive when the limit of |aₙ₊₁/aₙ| equals 1

If aₙ → 0 then Σaₙ converges

Applying the ratio test to Σ n!/nⁿ gives:

L = 1/e, so the series converges

L = 1, so the test is inconclusive

The step size h in a central-difference estimate of f′(x) is halved. If round-off is negligible, the truncation error becomes approximately:

One quarter of its previous value

Series Convergence, Experimental Data and Least Squares, and Numerical Differentiation

Four sub-items of Section 1 of the GATE Metallurgical Engineering (MT) paper that no shared mathematics chapter teaches, written here for this paper alone so that its Engineering Mathematics is complete. From the Calculus heading: "sequences and series; test for convergence". From Probability and Statistics: "analysis of experimental data; linear least squares method". From Numerical Methods: numerical "differentiation (Taylor series expansion)". The rest of MT’s mathematics — linear algebra, limits and derivatives, partial derivatives and maxima and minima, Fourier series, vector calculus, differential equations, probability and distributions, root finding and the trapezoidal rule — is in the shared chapters. The examples lean metallurgical where they can, because the least-squares line is how a metallurgist extracts an activation energy, a Hall-Petch constant or a parabolic rate constant from a table of measurements.

1. Sequences, series and the tests for convergence

A sequence aₙ converges to L if its terms get and stay arbitrarily close to L; a monotone sequence that is bounded converges. A series Σaₙ converges if its sequence of partial sums Sₙ = a₁ + ... + aₙ converges. Two families are the yardsticks for everything else. The geometric series Σrⁿ (n from 0) converges exactly when |r| < 1, to 1/(1 − r) — or a/(1 − r) with first term a. The p-series Σ1/nᵖ converges exactly when p > 1; with p = 1 it is the harmonic series, which diverges even though its terms go to zero.

The tests, and what each one can and cannot decide
Test	Statement	Watch out
nth-term (divergence) test	If aₙ does not tend to 0, Σaₙ diverges	aₙ → 0 is necessary, not sufficient: Σ1/n diverges
Comparison and limit comparison	For positive terms, if aₙ ≤ bₙ and Σbₙ converges, so does Σaₙ; if aₙ/bₙ → a finite non-zero limit, both behave alike	Compare with a p-series or a geometric series
Ratio test (d’Alembert)	L = lim \|aₙ₊₁/aₙ\|: converges if L < 1, diverges if L > 1	L = 1 decides nothing (every p-series gives L = 1)
Root test (Cauchy)	L = lim \|aₙ\|^(1/n), same conclusions	Best when aₙ is an nth power
Integral test	For positive decreasing f(n) = aₙ, Σaₙ and ∫f(x)dx converge or diverge together	This is how the p-series result is proved
Alternating series (Leibniz)	Σ(−1)ⁿbₙ converges if bₙ decreases to 0	May converge only conditionally

A series converges absolutely if Σ|aₙ| converges, and absolute convergence implies convergence. A series that converges but not absolutely converges conditionally: the alternating harmonic series 1 − 1/2 + 1/3 − ... converges (to ln 2) while Σ1/n diverges. For absolutely convergent series the terms may be rearranged freely; for conditionally convergent ones they may not.

2. Power series, the radius of convergence and Taylor series

A power series Σcₙ(x − a)ⁿ converges absolutely for |x − a| < R and diverges for |x − a| > R, where the radius of convergence follows from the ratio test: R = lim |cₙ/cₙ₊₁| (or 1/lim |cₙ|^(1/n)). The endpoints x = a ± R must be tested separately. Taylor’s theorem expands a smooth function about a: f(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + ..., the Maclaurin series when a = 0. The standard ones, with their radii: eˣ = Σxⁿ/n! (all x); sin x and cos x (all x); ln(1 + x) = x − x²/2 + x³/3 − ... (−1 < x ≤ 1); 1/(1 − x) = Σxⁿ (|x| < 1). Truncating a Taylor series leaves a remainder of the order of the first term dropped — the idea on which the next two sections both rest.

🧠 Factorials and powers decide the ratio test at a glance

n! beats any power cⁿ, and any power cⁿ with c > 1 beats any polynomial in n. So Σnᵏ/cⁿ converges for c > 1, Σcⁿ/n! converges for every c, and Σn!/cⁿ diverges. Σn!/nⁿ converges too: its ratio tends to 1/e.

3. Analysis of experimental data and the linear least squares method

Repeated measurements x₁ ... xₙ of one quantity are summarised by their mean x̄ = Σxᵢ/n and their sample standard deviation s = √(Σ(xᵢ − x̄)²/(n − 1)) — the divisor n − 1 because one degree of freedom has been used to find x̄. The uncertainty of the mean itself is the standard error s/√n, which falls only as the square root of the number of repeats. Random errors scatter results about the true value and are reduced by repetition; systematic errors shift every result the same way and are not. When independent quantities are combined, their uncertainties add in quadrature: for a sum or difference the absolute uncertainties, σ_z² = σ_x² + σ_y²; for a product or quotient the relative ones, (σ_z/z)² = (σ_x/x)² + (σ_y/y)².

To fit a straight line y = a + bx to n points (xᵢ, yᵢ), the least squares method chooses a and b to minimise the sum of squared vertical deviations S = Σ(yᵢ − a − bxᵢ)². Setting ∂S/∂a = 0 and ∂S/∂b = 0 gives the normal equations Σy = na + bΣx and Σxy = aΣx + bΣx², whose solution is b = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²) and a = ȳ − b x̄: the fitted line always passes through the centroid (x̄, ȳ). The correlation coefficient r = (nΣxy − ΣxΣy)/√((nΣx² − (Σx)²)(nΣy² − (Σy)²)) lies between −1 and +1 and measures how nearly linear the data are; r² is the fraction of the variance in y the line explains. A non-linear law is fitted by first linearising it, which is how metallurgical constants are found: ln k against 1/T is a line of slope −Q/R (Arrhenius); x² against t is a line of slope k_p (parabolic oxidation); σ_y against d^(−½) is a line of slope k_y and intercept σ₀ (Hall-Petch); log ε̇ against log σ is a line of slope n (creep).

A least-squares fit worked in full: (1, 2), (2, 3), (3, 5), (4, 6)
Quantity	Value
n, Σx, Σy	4, 10, 16
Σxy, Σx²	2 + 6 + 15 + 24 = 47; 1 + 4 + 9 + 16 = 30
Slope b	(4 × 47 − 10 × 16)/(4 × 30 − 10²) = 28/20 = 1.4
Intercept a	ȳ − b x̄ = 4 − 1.4 × 2.5 = 0.5
Correlation r	28/√(20 × 40) = 0.990

4. Numerical differentiation from the Taylor series

Every finite-difference formula for a derivative is read off Taylor expansions at neighbouring points, and the first term left out is its truncation error. From f(x + h) = f(x) + hf′(x) + h²f″(x)/2 + h³f‴(x)/6 + ..., the forward difference is f′(x) ≈ [f(x + h) − f(x)]/h with error −(h/2)f″ — first order, O(h); the backward difference [f(x) − f(x − h)]/h is also O(h). Subtracting the expansions for x + h and x − h cancels the even terms, giving the central difference f′(x) ≈ [f(x + h) − f(x − h)]/(2h) with error −(h²/6)f‴ — second order, O(h²), so halving h cuts the error fourfold. Adding them cancels the odd terms and gives the second derivative, f″(x) ≈ [f(x + h) − 2f(x) + f(x − h)]/h², also O(h²) with error −(h²/12)f⁗. A formula of order p is exact for polynomials of degree up to p: the central second difference is exact for any cubic.

Making h smaller does not improve the answer without limit. The truncation error falls as hᵖ, but the round-off error — the finite precision of f divided by a small h — grows as 1/h, so there is an optimum step below which the computed derivative gets worse. Measured data carry noise that plays the same role as round-off, which is why a derivative of experimental data (a cooling rate from a thermocouple trace, say) is taken from a smoothed or least-squares-fitted curve rather than from successive raw readings.

⚠️ The order is in h, not in the number of points

The forward and central formulas both use two function values, yet one is O(h) and the other O(h²). What decides the order is which Taylor terms cancel. A question that asks what happens to the error when h is halved is testing exactly this: it halves for forward and backward differences and quarters for central ones.

Key takeaways

Geometric Σrⁿ converges iff |r| < 1 (sum 1/(1 − r)); p-series Σ1/nᵖ iff p > 1. aₙ → 0 is necessary, not sufficient.
Ratio and root tests: L < 1 converges, L > 1 diverges, L = 1 undecided; the radius of a power series is lim |cₙ/cₙ₊₁|; absolute convergence implies convergence.
Sample standard deviation uses n − 1; the standard error of the mean is s/√n; uncertainties add in quadrature, absolute for sums and relative for products.
Least squares: b = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²), a = ȳ − b x̄, the line passes through (x̄, ȳ); linearise Arrhenius, parabolic and Hall-Petch laws before fitting.
Forward and backward differences are O(h); the central difference [f(x + h) − f(x − h)]/(2h) and the second difference [f(x + h) − 2f(x) + f(x − h)]/h² are O(h²).

Practice questions (12)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which of the following series converges?
1. Σ 1/n²
2. Σ 1/n
3. Σ 1/√n
4. Σ n/(n + 1)
Show answer
Answer: A — Σ 1/n²
Σ1/nᵖ converges only for p > 1, so Σ1/n² converges (to π²/6) while the harmonic series (p = 1) and Σ1/√n (p = ½) diverge. Σn/(n + 1) fails even the nth-term test, since its terms tend to 1, not 0.
What is the sum of the series 1 + 2/3 + (2/3)² + (2/3)³ + ... ?
Numerical answer — type the value.
Show answer
Answer: 3
A geometric series with first term 1 and ratio r = 2/3, |r| < 1, sums to 1/(1 − r) = 1/(1/3) = 3. Check with partial sums: 1, 1.667, 2.111, 2.407, ... creeping up to 3. Using a/(1 + r) gives 0.6.
What is the radius of convergence of the power series Σ n xⁿ/2ⁿ (n from 1)?
Numerical answer — type the value.
Show answer
Answer: 2
With cₙ = n/2ⁿ, |cₙ₊₁/cₙ| = ((n + 1)/n)(1/2) → 1/2, so R = 2. Root test agrees: |cₙ|^(1/n) = n^(1/n)/2 → 1/2. The factor n changes neither limit; at x = ±2 the terms are ±n, which do not tend to zero, so both endpoints diverge.
Which of the following statements about infinite series are correct?
1. If Σaₙ converges then aₙ → 0
2. The alternating harmonic series 1 − 1/2 + 1/3 − ... converges conditionally
3. The ratio test is inconclusive when the limit of |aₙ₊₁/aₙ| equals 1
4. If aₙ → 0 then Σaₙ converges
Show answer
Answer: A — If Σaₙ converges then aₙ → 0; B — The alternating harmonic series 1 − 1/2 + 1/3 − ... converges conditionally; C — The ratio test is inconclusive when the limit of |aₙ₊₁/aₙ| equals 1
A convergent series must have terms tending to zero (A). The alternating harmonic series converges by the Leibniz test but Σ1/n diverges, so it is conditionally convergent (B). L = 1 happens for both Σ1/n (divergent) and Σ1/n² (convergent), so it decides nothing (C). D is the converse of A and is false: the harmonic series is the counterexample.
Applying the ratio test to Σ n!/nⁿ gives:
1. L = 1/e, so the series converges
2. L = e, so the series diverges
3. L = 1, so the test is inconclusive
4. L = 0, so the series converges
Show answer
Answer: A — L = 1/e, so the series converges
aₙ₊₁/aₙ = (n + 1)! nⁿ/((n + 1)ⁿ⁺¹ n!) = (n/(n + 1))ⁿ = 1/(1 + 1/n)ⁿ → 1/e ≈ 0.368 < 1, so it converges. Reading (1 + 1/n)ⁿ → e the wrong way up gives L = e; forgetting that the nⁿ grows as fast as n! does gives L = 0.
Three repeated hardness readings are 10, 12 and 14 (arbitrary units). What is their sample standard deviation?
Numerical answer — type the value.
Show answer
Answer: 2
Mean = 12; deviations −2, 0, 2; Σd² = 8. With n − 1 = 2, s = √(8/2) = 2. The population formula with n = 3 gives √(8/3) = 1.63, which understates the scatter of so small a sample. The standard error of the mean is 2/√3 = 1.15.
A least-squares straight line y = a + bx is fitted to the points (1, 2), (2, 3), (3, 5) and (4, 6). What value of y does the fitted line predict at x = 5?
Numerical answer — type the value.
Show answer
Answer: 7.5
n = 4, Σx = 10, Σy = 16, Σxy = 47, Σx² = 30. b = (4 × 47 − 10 × 16)/(4 × 30 − 100) = 28/20 = 1.4 and a = ȳ − b x̄ = 4 − 1.4 × 2.5 = 0.5, so y(5) = 0.5 + 1.4 × 5 = 7.5. Check: the line passes through the centroid (2.5, 4), and its residuals 0.1, −0.3, 0.3, −0.1 sum to zero. Drawing a line through the end points (1, 2) and (4, 6) instead predicts 7.33.
Isothermal oxidation data (scale thickness x against time t) are to be tested for parabolic kinetics and the rate constant extracted by linear least squares. The quantity to plot on the vertical axis against t is:
1. x²
2. x
3. ln x
4. 1/x
Show answer
Answer: A — x²
The parabolic law x² = k_p t is linear in x² against t, with slope k_p and zero intercept, so least squares on (t, x²) gives k_p and the residuals show whether the law holds. x against t tests the linear law; ln x against ln t would give the exponent (2 for parabolic, as slope ½ of ln x on ln t) but not k_p directly; 1/x has no such law.
Using the forward difference with step h = 0.1, what is the estimate of the derivative of f(x) = x² at x = 1?
Numerical answer — type the value.
Show answer
Answer: 2.1
[f(1.1) − f(1)]/0.1 = (1.21 − 1)/0.1 = 2.1. The exact derivative is 2, and the error 0.1 is exactly (h/2)f″ = 0.05 × 2, the first term dropped from the Taylor series. The central difference [f(1.1) − f(0.9)]/0.2 = (1.21 − 0.81)/0.2 = 2 is exact for a quadratic.
Using the central difference with h = 0.1, estimate the derivative of f(x) = eˣ at x = 0 (to four decimal places).
Numerical answer — type the value.
Show answer
Answer: 1.0017
[e^0.1 − e^−0.1]/0.2 = (1.105171 − 0.904837)/0.2 = 0.200334/0.2 = 1.0017. The exact value is 1, and the error 0.00167 matches the predicted (h²/6)f‴(0) = 0.01/6 = 0.00167. The forward difference (e^0.1 − 1)/0.1 = 1.0517 is thirty times worse with the same h.
Using the central second-difference formula with h = 0.1, estimate the second derivative of f(x) = x³ at x = 1.
Numerical answer — type the value.
Show answer
Answer: 6
[f(1.1) − 2f(1) + f(0.9)]/h² = (1.331 − 2 + 0.729)/0.01 = 0.060/0.01 = 6. The exact value is f″(1) = 6x = 6: the formula’s error is −(h²/12)f⁗, and f⁗ = 0 for a cubic, so it is exact. Dividing by h instead of h² gives 0.6.
The step size h in a central-difference estimate of f′(x) is halved. If round-off is negligible, the truncation error becomes approximately:
1. One quarter of its previous value
2. One half of its previous value
3. Unchanged
4. Twice its previous value
Show answer
Answer: A — One quarter of its previous value
The central difference cancels the even Taylor terms and its error is −(h²/6)f‴, second order, so halving h divides the error by four. One half is the answer for the first-order forward and backward differences. If h were made very small, round-off would eventually make the error grow again.