Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 10

The comparison test for series states that if $0\le a_n\le b_n$ for all sufficiently large $n$, then convergence of $\sum b_n$ implies convergence of $\sum a_n$; conversely, divergence of $\sum a_n$ implies divergence of $\sum b_n$. A summable bound on $|a_n|$ proves absolute convergence of $\sum a_n$.

For the ratio test, suppose the terms are eventually nonzero and $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$ exists. If $L<1$, the series converges absolutely. If $L>1$, including $L=\infty$, the terms fail to tend to zero and the series diverges. If $L=1$, the test is inconclusive. An eventually zero series converges separately. Convergence means that the sequence of partial sums has a finite limit; the terms tending to zero is only a necessary condition for a convergent series.

For every real $\alpha$, $|\cos (\alpha n)/n^2|\le 1/n^2$. The bounding P-series with exponent two converges. The comparison test for series therefore givesOscillations of cosine are unnecessary for convergence here.

This is a geometric series with ratio $r=2\alpha /({\alpha }^2+1)$. Since ${\alpha }^2+1>0$, the condition $|r|<1$ is equivalent toIt holds for every real $\alpha$ except $\alpha =\pm 1$. At $\alpha =1$ the terms are all one; at $\alpha =-1$ they alternate between $-1$ and $1$. Neither sequence of terms tends to zero. ThusAll the convergent cases have absolute convergence, including $\alpha =0$, when every term is zero.

For $a_n=n^2{e}^{-\alpha n}$, the ratio test givesThis is less than one exactly when $\alpha >0$. For $\alpha =0$ the terms are $n^2$, and for $\alpha <0$ they grow exponentially; in both cases they fail to tend to zero. Consequently

Rationalize the difference to reveal its small size instead of comparing its two large summands separately:For any fixed real $\alpha$, the right-hand series is a constant multiple of a convergent P-series. The comparison test for series provesThe case $\alpha =0$ is the identically zero series. For nonzero $\alpha$, its summand is asymptotic to ${\alpha }^4/(2n^2)$, explaining the same convergence behavior as the inverse-square series.