Sequences and Infinite Series Questions and Answers

Answer: A

The series is convergent.

\(\sum 1/n^{p}\) is the p-series. By the integral test it behaves like \(\int_{1}^{\infty} x^{-p}\,dx\), which is finite exactly when \(p \gt 1\), so the series converges for \(p \gt 1\) and diverges for \(p \le 1\) (the harmonic series is the case \(p = 1\)).

It cannot be divergent here for that reason. 'Oscillatory' describes a series whose partial sums swing without settling, which needs terms of changing sign; every term here is positive, so the partial sums only increase. 'Asymptotic' is a property of expansions, not a convergence class of a series of positive terms.

Answer: C

The sequence is converging.

\(\frac{n}{n - 1} = 1 + \frac{1}{n - 1}\), which tends to 1 as \(n \to \infty\), so \(a_{n} = \left(\frac{n}{n - 1}\right)^{3} \to 1\). The terms decrease steadily toward this limit (\(a_{2} = 8\), \(a_{3} = 3.375\), \(a_{4} \approx 2.37\), ...), so the sequence is monotonic and convergent.

It is not oscillatory, since the terms never alternate, and not divergent, since it has a finite limit. For \(n \ge 2\) it is also bounded (between 1 and 8), but boundedness is a property every convergent sequence has; the key takes 'converging' as the precise description of its type.

Answer: B

The Maclaurin series is \(f(x) = f(0) + xf'(0) + \frac{x^{2}}{2!}f''(0) + \frac{x^{3}}{3!}f'''(0) + \cdots\).

Maclaurin’s series is the Taylor series expanded about \(x = 0\). The general Taylor series about a point \(a\) is

\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} \]

and with \(a = 0\) every term has the derivative evaluated at 0 and a power of \(x\), all with plus signs.

The options evaluating derivatives at 1 are not Maclaurin series, and a Taylor series about 1 would need powers of \((x - 1)\), not \(x\). The alternating-sign version at 0 is the series for \(f(-x)\), not \(f(x)\).

Answer: A

The sum is \(\frac{1}{1 - q} - \frac{q^{r+1}}{1 - q}\).

This is a finite geometric series of \(r + 1\) terms with first term 1 and ratio \(q\) (for \(q \neq 1\)). Let \(S = 1 + q + \cdots + q^{r}\). Then

\[ \begin{aligned} S - qS &= 1 - q^{r+1} \\ S &= \frac{1 - q^{r+1}}{1 - q} = \frac{1}{1 - q} - \frac{q^{r+1}}{1 - q} \end{aligned} \]

The option with a plus sign between the two fractions gets the sign of the second term wrong. Written over \(q - 1\), the correct sum is \(\frac{q^{r+1} - 1}{q - 1}\), so '\(\frac{1}{q - 1} + \frac{q^{r+1}}{q - 1}\)' also has the wrong sign on the 1. \(\frac{1}{q - 1}\) alone is the negative of the infinite sum \(\frac{1}{1 - q}\), which in any case applies only when \(|q| \lt 1\) and \(r\) grows without limit.

Answer: C

The series can be used for \(|x| \lt 1\).

\(1 + x + x^{2} + x^{3} + \ldots\) is a geometric series with common ratio x. It converges, and its sum equals \(1/(1 - x)\), exactly when \(|x| \lt 1\); its radius of convergence is 1, the distance from 0 to the singularity at \(x = 1\).

\(|x| \neq 1\) wrongly includes values such as x = 2, where the terms grow without bound. For \(x \lt -1\) the series also diverges. \(-1 \le x \lt 1\) wrongly includes \(x = -1\), where the series becomes 1 - 1 + 1 - 1 ..., which oscillates and has no sum, even though f(-1) = 1/2.

Answer: D

Series (P) diverges and series (Q) converges.

(P): the terms do not tend to zero. Since \(\ln(n\log n)/n \to 0\), \((n\log n)^{1/n} \to 1\), so each term tends to 1 and the series diverges by the n-th term test.

(Q): apply the ratio test:

\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \frac{(n+1)^{n+1}}{n^{n}}\cdot\frac{1}{(2n+2)(2n+1)} \\ &= \left(1+\frac{1}{n}\right)^{n}\frac{n+1}{(2n+2)(2n+1)} \to e \cdot 0 = 0 \end{aligned} \]

The ratio tends to 0, so (Q) converges: the factorial in the denominator grows much faster than \(n^{n}\). The other options misjudge one of the two series.

Answer: C

The correct matching is P → III, Q → I, R → II.

• P: the series telescopes, \(\frac{1}{(n+1)(n+2)} = \frac{1}{n+1}-\frac{1}{n+2}\), so the sum from n = 1 is \(\frac{1}{2}\) (III).
• Q: \(\int_{0}^{x}\sin t\,dt = 1-\cos x \approx \frac{x^{2}}{2}\) for small x, so the limit is \(3 \times \frac{1}{2} = \frac{3}{2}\) (I).
• R: the function is already a Fourier series with \(a_{1} = -\frac{1}{2}\), \(b_{1} = \frac{1}{2}\), \(b_{2} = \frac{1}{\sqrt{2}}\) and every other coefficient zero, so the sum of squares is \(\frac{1}{4}+\frac{1}{4}+\frac{1}{2} = 1\) (II).

A common slip in P is to start the telescoping at n = 0, which gives 1; the options pairing P with I or II follow from errors of that kind.