Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 9E (Analysis I)

The ratio test says that a series $\sum u_n$ whose terms are eventually nonzero converges absolutely if $\mathrm{limsup}|u_{n+1}/u_n|<1$; in particular, it suffices to have $|u_{n+1}/u_n|\le q<1$ eventually, since then its terms are bounded by a convergent geometric series. The comparison test for series states that, for nonnegative terms, comparison with a larger convergent series gives convergence, and comparison with a smaller divergent series gives divergence. For $u_n=n!/n^n$,By the binomial theorem, $(1+1/n{)}^n\ge 1+n(1/n)=2$. Thus $u_{n+1}/u_n\le 1/2$ for every $n\ge 1$, and induction gives $0<u_n\le 2^{1-n}$. The comparison test for series with $\sum 2^{1-n}=2$ proves

The limit comparison test says that if $u_n,v_n>0$ and $u_n/v_n\to L$ with $0<L<\infty$, then their series either both converge or both diverge. This follows because, eventually, $(L/2)v_n\le u_n\le (3L/2)v_n$, and the comparison test for series applies in both directions.

Here comparison with the harmonic series givesFor completeness, the natural logarithm grows slowly enough because for $x\ge 1$,so $(\log n{)}^2/n\le 16/\sqrt{n}\to 0$. The harmonic series diverges: every block from $n=2^k$ to $2^{k+1}-1$ has $2^k$ terms at least $2^{-k-1}$ and therefore contributes at least $1/2$. Hence

The alternating series test states that if $b_n\ge 0$ decreases to zero, then $\sum (-1{)}^n{b}_n$ converges. Indeed, its even and odd partial sums are monotone in opposite directions and bound each other, while their difference tends to zero. Here $b_n=1/(1+\sqrt{n})$ decreases to zero, so the series converges.

It fails absolute convergence, since for $n\ge 1$,The p-series $\sum n^{-1/2}$ diverges. This also follows by comparison with the divergent harmonic series, because $n^{-1/2}\ge n^{-1}$. The comparison test for series therefore shows that $\sum b_n$ diverges. Thus

For $s=\pm 1$, the identity $s/(2+s)=-1/3+(2/3)s$ givesThe alternating series test applies to $\sum (-1{)}^n/n$ because $1/n$ decreases to zero; its partial sums consequently have a finite limit. The harmonic series $\sum 1/n$ diverges to $+\infty$. Applying the displayed identity to each finite partial sum yieldsTherefore $\boxed{\text{the series diverges to }-\infty }$. Alternating signs alone do not satisfy the alternating series test: the magnitudes jump from $1/(3n)$ at even $n$ to $1/(n+1)$ at the next odd index.