Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 19

The Big O notation means that there exist constants $C>0$ and $\delta >0$ such thatwithin the domain of $f$. Thus $f(x)/x^2$ stays bounded in a punctured neighbourhood of zero. It need not have a limit: $f(x)=x^2\sin (1/x)$ is one example. The statement imposes no condition on a separately assigned value $f(0)$.

Use the Maclaurin series on the positive side:In the last expression the original square root covers $e^x-1$, whose leading term is $x>0$. Hence $g/f\to 0$ and $f/h\to 0$. All three are positive for sufficiently small positive $x$, soThus $h$ is greatest and $g$ least in the stated limit.

Apply the ratio test to the infinite series whose partial sums are $S_N$. Assuming the terms are eventually nonzero and the limit exists, letIf $\ell <1$, the series has absolute convergence, by eventual comparison with a decaying geometric series. If $\ell >1$, including an infinite limit, the terms fail to tend to zero and the series diverges. If $\ell =1$, the test is inconclusive: both $\sum 1/k$ and $\sum 1/k^2$ have that ratio limit, but only the second converges. A finite partial sum itself requires no convergence test; the issue is the limit as $N\to \infty$.

For $p\le 0$, the terms $k^{-p}$ do not tend to zero, so the series diverges. For $p>0$, group the positive terms into the dyadic blocks $2^j\le k<2^{j+1}$. Each block contains $2^j$ terms, and monotonicity givesWhen $p>1$, the upper bounds form a convergent geometric series with ratio $2^{1-p}<1$, proving convergence. When $0<p\le 1$, the lower bounds do not tend to zero, so infinitely many blocks each make a nonvanishing contribution and the sum diverges. Thus the p-series criterion isThis dyadic comparison is the mechanism of the Cauchy condensation test. In particular $p=1$ is the divergent harmonic series.