Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 2

The comparison test for series says that $0\le v_n\le C{w}_n$ eventually and convergence of $\sum w_n$ imply convergence of $\sum v_n$. Conversely, an eventual positive lower comparison with a divergent positive series proves divergence. The ratio test uses $L={\lim }_{n\to \infty }{v}_{n+1}/v_n$: $L<1$ proves convergence, $L>1$ proves divergence, and $L=1$ is inconclusive. Removing finitely many initial zero terms does not affect these conclusions.

For $n\ge 1$, $\ln n\le n$ and $2n^3+1\ge 2n^3$. Hence $0\le \ln n/(2n^3+1)\le 1/(2n^2)$. The comparison p-series has exponent two and converges, so the series converges by the comparison test.

The summand $n/(n+4)$ tends to one rather than zero. The term test for divergence therefore proves the series diverges. A ratio limit equal to one would be inconclusive and is unnecessary here.

The elementary bound $|\sin t|\le |t|$ gives $0\le {\sin }^2(1/n)\le 1/n^2$. Comparison with a convergent p-series proves the series converges.

For $n\ge 1$, the ratio of positive terms isThe ratio test proves the series converges. Its zero term at $n=0$ has no effect.