Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics II / 2

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges when its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence.

The comparison test says that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Indeed, the partial sums of the nonnegative terms $a_n$ are increasing, and are bounded above by a fixed initial contribution plus the sum of the comparison series. The bounded monotone sequence theorem proves their convergence. Finite initial terms do not affect convergence. Applying this result to $|a_n|$ also proves absolute convergence whenever $|a_n|\le b_n$ eventually with $\sum b_n<\infty$.

The implication is true, with absolute convergence. Since $n^{3/2}{a}_n\to 0$, there is $N$ such that $|a_n|\le n^{-3/2}$ for $n\ge N$. The p-series $\sum n^{-3/2}$ converges because its exponent exceeds one. The comparison test proves $\sum |a_n|<\infty$, and absolute convergence implies convergence of $\sum a_n$.

The implication is false. SetThen $n{a}_n=1/\log (n+1)\to 0$. For $n\ge 2$, $\log (n+1)\le \log (2n)\le 2\log n$, so $a_n\ge 1/(2n\log n)$. The integral test givesThus the comparison test shows that $\sum a_n$ diverges.

The binomial theorem gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)=4^n$. ThereforeThe geometric series with ratio $4/9$ converges. By the comparison test, the given series is absolutely convergent.

For $t\ge 0$, $|\sin t|\le t$, soThe integral test applies to the decreasing positive comparison function, andThe comparison test proves that the series converges; since all terms are nonnegative, this is also absolute convergence.

Let $u(x)=(\log x)x^{-2/3}$. Its derivative iswhich is negative for $x>e^{3/2}$. Also $u(x)\to 0$, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of $\sum (-1{)}^{n}u(n)$.

For $n\ge 3$, $u(n)\ge n^{-2/3}$, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.