Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 19

For a series $\sum u_k$ with eventually nonzero terms, supposeexists, allowing $L=+\infty$. The ratio test says that if $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. In the divergent case the terms fail to tend to zero. For example, the P-series with exponents $1$ and $2$ both have ratio limit $1$ but have different convergence behaviour.

The term at $k=0$ is finite. For every $k\ge 1$,The comparison P-series $\sum k^{-2}$ converges, so the comparison test for series proves that the given series converges absolutely.

For $u_k=a^{2k+1}/(2k+1)>0$, the consecutive ratio isThus the ratio test proves convergence when $0<a<1$ and divergence when $a>1$. At $a=1$, the series is ${\sum }_{k\ge 0}1/(2k+1)$, which diverges by comparison with $\frac{1}{2}{\sum }_{k\ge 0}1/(k+1)$. ThereforeAll terms are nonnegative, so convergence here is also absolute convergence.

L'Hôpital's rule applies to a quotient $f(x)/g(x)$ of differentiable functions near a limiting point, excluding that point itself, provided $g^{\prime }(x)\ne 0$ there. If the quotient has an indeterminate form $0/0$ or $\infty /\infty$, and the limit of $f^{\prime }(x)/g^{\prime }(x)$ exists as a finite or infinite value $L$, thenThe rule also applies to one-sided limits and to limits at infinity, under the corresponding hypotheses. Existence of the derivative-ratio limit and the indeterminate-form hypotheses are required; an arbitrary quotient cannot be replaced by a quotient of derivatives.

Write the product as a quotient with indeterminate form $(-\infty )/(+\infty )$. By L'Hôpital's rule,

The original PDF specifies $x\to a$. The quotient is the difference quotient for the function $f(x)=x^x=e^{x\ln x}$ at the positive point $a$. The chain rule gives $f^{\prime }(x)=x^x(\ln x+1)$, so

Use the root limit of a finite sum of powers. Put $M=\max \{1,a,a^2/2\}\ge 1$. For every $x>0$, at least one of the three terms is $M^x$ and none is larger, soTaking the positive $x$th root givesSince $3^{1/x}\to 1$, the squeeze theorem givesIndeed, for $a\le 1$ both remaining bases are at most $1$; for $1\le a\le 2$, $a\ge a^2/2$; and for $a\ge 2$, $a^2/2\ge a$. The endpoint cases $a=0,1,2$ are included in the same bound.