Series Convergence, Limits in Two Variables, Correlation–Regression and Bayesian Inference
1. Sequences, series and the first two questions to ask
A sequence aₙ converges to L if its terms get and stay arbitrarily close to L as n → ∞; aₙ = n/(n + 1) converges to 1, aₙ = (−1)ⁿ oscillates and has no limit. A series Σaₙ is the sequence of its partial sums Sₙ = a₁ + … + aₙ, and the series converges when Sₙ does. Two questions settle most GATE items before any named test is needed. First, do the terms go to zero? If aₙ does not tend to 0 the series diverges — this is the nth-term (divergence) test, and it can only ever prove divergence. Second, is it a benchmark series? The geometric series Σ rⁿ converges to a/(1 − r) exactly when |r| < 1, and the p-series Σ 1/nᵖ converges exactly when p > 1. The harmonic series Σ 1/n (p = 1) diverges even though its terms go to zero.
2. Tests for convergence, and which one the series is asking for
| Test | Statement | Reach for it when |
|---|---|---|
| Comparison | 0 ≤ aₙ ≤ bₙ: Σbₙ converges ⇒ Σaₙ converges; Σaₙ diverges ⇒ Σbₙ diverges | The term is bounded by a p-series or geometric term |
| Limit comparison | If aₙ/bₙ → c with 0 < c < ∞, both converge or both diverge | Rational functions of n: compare with the ratio of leading powers |
| Ratio (d’Alembert) | L = lim |aₙ₊₁/aₙ|: L < 1 converges, L > 1 diverges, L = 1 no conclusion | Factorials and powers such as n!, 2ⁿ, nᵏ |
| Root (Cauchy) | L = lim |aₙ|^(1/n): same three verdicts as the ratio test | The whole term is raised to the power n |
| Integral | f positive, decreasing, f(n) = aₙ: Σaₙ and ∫₁^∞ f(x) dx converge together | Terms like 1/(n ln n); it is also how the p-series result is proved |
| Leibniz (alternating) | Σ(−1)ⁿ⁺¹bₙ converges if bₙ decreases and bₙ → 0 | Signs alternate; then check absolute convergence separately |
Absolute against conditional convergence. A series converges absolutely if Σ|aₙ| converges, and absolute convergence implies convergence. Σ(−1)ⁿ⁺¹/n² converges absolutely. Σ(−1)ⁿ⁺¹/n and Σ(−1)ⁿ⁺¹/√n converge by Leibniz, but their absolute series are p-series with p ≤ 1 and diverge — so they converge conditionally. Worked ratio test: for Σ n²/2ⁿ, aₙ₊₁/aₙ = (n + 1)²/(2n²) → 1/2 < 1, so the series converges. For Σ n!/nⁿ the ratio is (n/(n + 1))ⁿ → 1/e < 1, so it converges too.
3. Limits, continuity and differentiability in two variables
In one variable a point can be approached from two sides; in two variables it can be approached along infinitely many paths, and the limit exists only if every path gives the same value. That makes disproof easy and proof harder. To show a limit does not exist, find two paths that disagree. For f(x, y) = xy/(x² + y²) at the origin, the path y = mx gives m/(1 + m²), which depends on m — 0 along an axis, 1/2 along y = x — so the limit does not exist. To show a limit does exist, bound it: for x²y/(x² + y²), |x²y/(x² + y²)| ≤ |y| → 0, so the limit is 0. Polar substitution x = r cos θ, y = r sin θ does the same job: if the expression tends to a value as r → 0 uniformly in θ, that is the limit.
- Continuity at (a, b) means the limit exists and equals f(a, b). A function defined piecewise with f(0, 0) = 0 is continuous there only if the limit at the origin is also 0.
- Partial derivatives existing does not imply continuity. xy/(x² + y²) with f(0, 0) = 0 has fₓ(0, 0) = f_y(0, 0) = 0, because f vanishes along both axes, yet it is not continuous at the origin.
- Differentiability implies continuity, and continuous first partial derivatives in a neighbourhood are enough to guarantee differentiability. The chain of implications runs one way only.
4. Correlation and regression analysis
For paired data (xᵢ, yᵢ), the covariance is cov(x, y) = (1/n)Σ(xᵢ − x̄)(yᵢ − ȳ), and Karl Pearson’s correlation coefficient is r = cov(x, y)/(σₓσᵧ), which always lies in [−1, 1]. It measures linear association only: r = 0 means no linear relation, not independence — y = x² on symmetric x values has r = 0 and perfect dependence. The least-squares regression line of y on x is y − ȳ = bᵧₓ(x − x̄) with bᵧₓ = cov/σₓ² = r σᵧ/σₓ, and the line of x on y is x − x̄ = bₓᵧ(y − ȳ) with bₓᵧ = r σₓ/σᵧ. Multiply the two coefficients and the standard deviations cancel: r² = bᵧₓ · bₓᵧ.
| Property | Consequence |
|---|---|
| Both regression lines pass through (x̄, ȳ) | Solving the two line equations simultaneously gives the means |
| bᵧₓ, bₓᵧ and r share one sign | r = ±√(bᵧₓbₓᵧ), taking the common sign of the coefficients |
| bᵧₓ · bₓᵧ = r² ≤ 1 | Decides which given line is y on x: the wrong assignment gives a product above 1 |
| r = 0 | The lines become y = ȳ and x = x̄, perpendicular to each other |
| r = ±1 | The two lines coincide |
| Change of origin and scale | r is unchanged (up to sign if one scale factor is negative); regression coefficients change with scale |
5. Bayesian statistics — prior, likelihood and posterior
Bayesian statistics treats an unknown — a disease state, a parameter θ — as uncertain, describes that uncertainty with a prior probability, and updates it with data through Bayes’ theorem: posterior ∝ likelihood × prior, P(θ | data) = P(data | θ)P(θ)/P(data). The denominator P(data) is the total probability of the evidence summed over every possibility, and forgetting it — reporting likelihood × prior unnormalised, or reporting the likelihood itself — is the standard error. Today’s posterior is tomorrow’s prior, so evidence arriving in two batches gives the same answer as all of it at once.
Conjugate updating. When a success probability θ has a Beta(a, b) prior and n trials give k successes (a binomial likelihood), the posterior is Beta(a + k, b + n − k) — the prior behaves like a earlier successes and b earlier failures. The posterior mean is (a + k)/(a + b + n), which always lies between the prior mean a/(a + b) and the sample proportion k/n, and moves toward k/n as n grows. With a Beta(2, 2) prior and 7 successes in 10 trials the posterior is Beta(9, 5) with mean 9/14 ≈ 0.64, between the prior 0.5 and the data 0.7. The frequentist estimate is just k/n; the Bayesian one carries the prior and yields a whole distribution for θ, from which a credible interval is read directly.
Key takeaways
- aₙ → 0 is necessary for convergence and never sufficient; Σ1/nᵖ converges only for p > 1, and Σrⁿ only for |r| < 1.
- A ratio or root limit of 1 is inconclusive — fall back on comparison or the integral test; alternating series need Leibniz and then a separate check for absolute convergence.
- A two-variable limit exists only if every path agrees; two disagreeing paths, even a line and a parabola, disprove it.
- r² = bᵧₓbₓᵧ, all three share a sign, and both regression lines meet at (x̄, ȳ); r = 0 is no linear relation, not independence.
- Posterior ∝ likelihood × prior, normalised by the total probability of the evidence; a Beta(a, b) prior with k successes in n becomes Beta(a + k, b + n − k).
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.
The series Σ (n = 1 to ∞) 1/nᵖ converges if and only if
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Answer: A — p > 1
The integral test compares Σ1/nᵖ with ∫₁^∞ x⁻ᵖ dx, which is finite exactly when p > 1. The boundary case p = 1 is the harmonic series, which diverges, so p ≥ 1 is the tempting wrong choice; p > 0 only makes the terms tend to zero, which is not enough.Applying the ratio test to Σ (n = 1 to ∞) n²/2ⁿ gives the limit L and the verdict
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Answer: A — L = 1/2, the series converges
aₙ₊₁/aₙ = [(n + 1)²/2ⁿ⁺¹]·[2ⁿ/n²] = (1/2)(1 + 1/n)² → 1/2 < 1, so the series converges. The polynomial factor n² only contributes (1 + 1/n)² → 1; it is the 2ⁿ that decides. Inverting the ratio gives the wrong L = 2.Which of the following series converge?
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Answer: A — Σ 1/n^1.5; C — Σ (−1)ⁿ⁺¹/n
Σ1/n^1.5 is a p-series with p = 1.5 > 1 and converges. Σ(−1)ⁿ⁺¹/n converges by Leibniz (1/n decreases to 0), though only conditionally. Σ1/√n is a p-series with p = 1/2 and diverges despite its terms tending to zero, and Σ n/(n + 1) fails the nth-term test outright because its terms tend to 1.The sum of the series Σ (n = 0 to ∞) (2/3)ⁿ is ______.
Numerical answer — type the value.
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Answer: 3
A geometric series with first term 1 and ratio r = 2/3; since |r| < 1 it converges to 1/(1 − 2/3) = 3. Starting the index at n = 1 would give 2 instead — read the lower limit before using a/(1 − r).The series Σ (n = 1 to ∞) (−1)ⁿ⁺¹/√n is
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Answer: B — conditionally convergent
1/√n decreases to zero, so the Leibniz test gives convergence. The series of absolute values is Σ1/√n, a p-series with p = 1/2 ≤ 1, which diverges — so the convergence is conditional, not absolute. Choosing 'absolutely convergent' skips the second check.For f(x, y) = x²y/(x⁴ + y²), the limit of f as (x, y) → (0, 0)
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Answer: C — does not exist
Along y = mx the function is mx/(x² + m²) → 0 for every m, which is the trap. Along the parabola y = x² it is x⁴/(2x⁴) = 1/2. Two paths give different values, so the limit does not exist; agreement along all straight lines is not enough.For a bivariate data set the regression coefficient of y on x is 0.8 and of x on y is 0.45. The correlation coefficient r is ______.
Numerical answer — type the value.
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Answer: 0.6
r² = bᵧₓ·bₓᵧ = 0.8 × 0.45 = 0.36, so |r| = 0.6, and r is positive because both coefficients are positive. Averaging the two coefficients (0.625) is the tempting wrong move; r is their geometric mean, not the arithmetic one.The two regression lines of a bivariate distribution are 3x + 2y = 26 and 6x + y = 31. The correlation coefficient between x and y is ______.
Numerical answer — type the value.
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Answer: -0.5
Taking 3x + 2y = 26 as y on x gives bᵧₓ = −3/2; taking 6x + y = 31 as x on y gives bₓᵧ = −1/6. Their product is 1/4 ≤ 1, so this assignment is the valid one and r = −√(1/4) = −0.5, negative like both coefficients. The other assignment gives a product of 4, impossible for r². Reporting +0.5 forgets the sign.Which of the following statements about correlation and linear regression are correct?
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Answer: A — |r| is the geometric mean of the two regression coefficients.; B — If r = 0, the two regression lines are perpendicular.
r² = bᵧₓbₓᵧ makes |r| their geometric mean, and at r = 0 the lines reduce to y = ȳ and x = x̄, which are perpendicular. Both coefficients equal r times a positive ratio of standard deviations, so they always share r’s sign. And r = 0 rules out only a linear relation — y = x² on symmetric data has r = 0 with complete dependence.A condition has prevalence 1%. A test for it has sensitivity 95% and specificity 90%. The probability that a person who tests positive has the condition is ______ (round off to three decimal places).
Numerical answer — type the value.
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Answer: 0.088
P(+) = 0.95 × 0.01 + 0.10 × 0.99 = 0.1085, the false-positive rate being 1 − specificity. P(D | +) = 0.0095/0.1085 = 0.0876 ≈ 0.088. Answering 0.95 confuses P(+ | D) with P(D | +), the error Bayes’ theorem exists to prevent.The probability θ that a cloned colony expresses a protein has a Beta(2, 2) prior. Of 10 colonies screened, 7 express it. The posterior mean of θ is ______ (round off to two decimal places).
Numerical answer — type the value.
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Answer: 0.64
Beta prior with binomial data is conjugate: the posterior is Beta(2 + 7, 2 + 3) = Beta(9, 5), with mean 9/(9 + 5) = 9/14 ≈ 0.64. It sits between the prior mean 0.5 and the sample proportion 0.7; reporting 0.7 ignores the prior entirely.The two lines of regression of a bivariate data set always intersect at
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Answer: A — the point of means (x̄, ȳ)
Each least-squares line is written as y − ȳ = bᵧₓ(x − x̄) or x − x̄ = bₓᵧ(y − ȳ), and both are satisfied at x = x̄, y = ȳ. That is why solving the two given line equations simultaneously is the quickest way to find the means; the origin works only for centred data.