Real Analysis Questions and Answers

Answer: B

The limit is 1.

Average the three terms of \(P(k\sqrt{2})\) separately. The constant 1 averages to 1. For the others write them as \(e^{2\pi i k\theta}\):

\[ \begin{aligned} e^{2\pi i k\sqrt{2}}&:\ \theta = \sqrt{2} \\ e^{3\pi i k\sqrt{2}}&:\ \theta = \tfrac{3}{2}\sqrt{2} \end{aligned} \]

Both values of \(\theta\) are irrational. By Weyl's equidistribution theorem, for irrational \(\theta\) the sequence \(k\theta\) mod 1 is equidistributed, so the averages of \(e^{2\pi ik\theta}\) tend to \(\int_{0}^{1}e^{2\pi it}dt = 0\). (Directly, the geometric sum is bounded while N grows.)

So the limit is 1 + 0 + 2 × 0 = 1. 4 would be P(0), the value if the exponentials stayed at 1, and 3 or 0 drop or keep the wrong terms.

Answer: C

The limit is 1.5.

Divide the numerator and denominator by \(n^{2}\), the highest power:

\[ \frac{3 + 5/n + 4/n^{2}}{4/n^{2} + 2} \to \frac{3}{2} = 1.5 \]

For a ratio of polynomials of equal degree, the limit is the ratio of the leading coefficients. 0 would need the denominator to be of higher degree. 0.75 is \(\frac{3}{4}\), dividing by the constant term 4 instead of the coefficient 2. 3 ignores the denominator's coefficient.

Answer: D

\((x_n)\) is not convergent and \((y_n)\) is convergent.

For even \(n\), \((-1)^{n} = 1\) and \(x_n = \frac{1}{n} \to 0\). For odd \(n\), \((-1)^{n} = -1\) and \(x_n = 2 + \frac{1}{n} \to 2\). Two subsequences with different limits mean \((x_n)\) does not converge.

For \((y_n)\), regroup the power so the standard limit \(\left(1 + \frac{1}{m}\right)^{m} \to e\) appears with \(m = 2n\):

\[ \begin{aligned} y_n &= \left[\left(1 + \frac{1}{2n}\right)^{2n}\right]^{3/2} \\ &\to e^{3/2} \end{aligned} \]

So \((y_n)\) converges. The options saying \((x_n)\) converges miss the alternating \((-1)^{n}\) term, which does not die out; the option saying \((y_n)\) diverges treats it like \((1 + \frac{1}{2n})^{n^{2}}\), where the exponent grows too fast.

Answer: D

The FALSE statement is the last one: for all \(\epsilon \gt 0\) and \(M \gt 0\) there is \(N \gt M\) with \(|x_N - 1| \gt \epsilon\).

Since \(x_n = (-1)^{n}\), the distance \(|x_n - 1|\) is 0 for even \(n\) and 2 for odd \(n\). It never exceeds 2, so for \(\epsilon = 2\) (or anything larger) no \(N\) satisfies \(|x_N - 1| \gt \epsilon\). A statement claimed for all \(\epsilon\) fails.

The other three are true. The first holds with \(\epsilon = 3\), since every distance is at most 2. The second holds with \(\epsilon = 1\): beyond any \(M\) there is an odd \(N\), where the distance is \(2 \gt 1\). The third holds because beyond any \(M\) there is an even \(N\), where the distance is \(0 \lt \epsilon\). Together, the second and third say that 1 is a limit point of the sequence but not its limit.

Answer: C

The TRUE statement is: if \(\sum a_n\) converges but \(\sum |a_n|\) does not, then \(\sum b_n\) does not converge.

The sequence \(b_n\) keeps the non-negative terms and \(c_n\) the negative ones, so

\[ \begin{aligned} b_n &= \tfrac{1}{2}(|a_n| + a_n) \\ |a_n| &= 2b_n - a_n \end{aligned} \]

If \(\sum a_n\) and \(\sum b_n\) both converged, \(\sum |a_n|\) would converge too. So when \(\sum a_n\) converges conditionally, \(\sum b_n\) must diverge, which also rules out the option saying it converges. The same argument with \(c_n = \tfrac{1}{2}(a_n - |a_n|)\) shows \(\sum c_n\) diverges as well, so the last option fails for the conditionally convergent series \(\sum \frac{(-1)^{n+1}}{n}\).

The first option fails for \(a_n = -1\): \(\sum a_n\) diverges, yet every \(b_n = 0\), so \(\sum b_n\) converges.

Answer: C

The pair is \(a = \frac{-1}{2}\) and \(b = \frac{1}{4}\).

Write \(t = \frac{3x}{1-x}\), so the term is \(\frac{t^{k}}{2^{\sqrt{k}}}\). Since \((2^{\sqrt{k}})^{1/k} = 2^{1/\sqrt{k}} \to 1\), the root test gives convergence for \(|t| \lt 1\) and divergence for \(|t| \gt 1\). At \(|t| = 1\) the terms are \(\pm 2^{-\sqrt{k}}\), which are summable because \(2^{\sqrt{k}}\) eventually exceeds \(k^{2}\). So the series converges exactly when \(|3x| \leq |1 - x|\):

\[ \begin{aligned} 9x^{2} &\leq (1-x)^{2} \\ 8x^{2} + 2x - 1 &\leq 0 \\ (4x - 1)(2x + 1) &\leq 0 \\ -\tfrac{1}{2} \leq x &\leq \tfrac{1}{4} \end{aligned} \]

The interval \((a,b)\) must sit inside \([-\frac{1}{2}, \frac{1}{4}]\), and \((-\frac{1}{2}, \frac{1}{4})\) does. The others reach outside it: \(b = \frac{1}{2}\) and \(b = \frac{1}{3}\) include \(x = 0.3\), where \(t = \frac{9}{7} \gt 1\); \(a = -1\) includes \(x = -0.9\), where \(|t| = \frac{27}{19} \gt 1\).

Answer: A

Both \((x_n)\) and \((y_n)\) are convergent.

For \((x_n)\), compare each term \(\frac{1}{k}\) with \(\int_{k}^{k+1}\frac{dt}{t}\). Since \(\frac{1}{t}\) is decreasing,

\[ \begin{aligned} x_n - x_{n+1} &= \log_{e}\tfrac{n+1}{n} - \tfrac{1}{n+1} \gt 0 \\ x_n &\geq \log_{e}(n+1) - \log_{e} n \gt 0 \end{aligned} \]

A decreasing sequence bounded below converges; its limit is Euler's constant, about 0.5772. The harmonic sum on its own diverges, but subtracting \(\log_{e} n\) cancels exactly that growth.

For \((y_n)\), \(\left|\frac{\cos t}{t^{2}}\right| \leq \frac{1}{t^{2}}\) and \(\int_{1}^{\infty}\frac{dt}{t^{2}} = 1\), so \(\int_{1}^{\infty}\frac{\cos t}{t^{2}}\,dt\) converges absolutely and \(y_n\) tends to it. The oscillation of \(\cos t\) would matter only with a slowly decaying denominator such as \(t\) or \(\sqrt{t}\). Every option saying either sequence fails to converge is therefore wrong.

Answer: B

The radius of convergence is \(R = \frac{9}{4}\), so \(4R = 9\).

Use the ratio test on the coefficients \(a_{n}\), the squared fractions. Going from \(n\) to \(n+1\) adds one factor \(2n+2\) on top and \(3n+2\) underneath:

\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \left(\frac{2n+2}{3n+2}\right)^{2} \\ &\to \left(\frac{2}{3}\right)^{2} = \frac{4}{9} \end{aligned} \]

so \(R = \lim \frac{a_{n}}{a_{n+1}} = \frac{9}{4}\). The option \(9R = 4\) gives \(R = \frac{4}{9}\), the reciprocal, which is the limit of the ratio itself rather than of its inverse. \(R = 1\) and \(R = \frac{1}{2}\) come from forgetting the square or misreading the factors.

Answer: B

P is NOT correct and Q is correct.

P fails because \(g(n)\) looks only at even arguments. \(\lim g(n)\) is the limit superior of the subsequence \(f(2), f(4), f(6), \ldots\), which can be smaller than \(\limsup f(n)\). For example, let \(f\) take the values 1 to 100 once each on \(n = 1, \ldots, 100\), then equal 1 at every even \(n\) and 100 at every odd \(n\) after that. It is onto, \(\limsup f(n) = 100\), but \(g(n) = 1\) for large \(n\).

Q holds. \(h(n)\) is the infimum of the tail from \(n\) onwards, so \(h\) is non-decreasing and bounded, and \(h(n) \to \liminf f(n)\) by definition. \(h(4n+1)\) is a subsequence of a convergent sequence, so it has the same limit.

Answer: C

The sequence has two limit points, 1 and -1.

The term \(6/n^{2}\) tends to 0, so the behaviour of \(a_{n}\) is decided by \((-1)^{n}\). Along even \(n\), \(a_{n} = 1 + 6/n^{2} \to 1\); along odd \(n\), \(a_{n} = -1 + 6/n^{2} \to -1\). Every subsequence eventually lies near one of these two values, so they are the only limit points.

'No limit point' is impossible for a bounded sequence (Bolzano-Weierstrass). 'One limit point' would need the sequence to converge, which it does not, since its even and odd terms approach different values. 'More than two' would need a third cluster value, and there is none.