Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 9F (Analysis I)

Put $s=\sin x$, so $-1\le s\le 1$. The printed summand is $(3s^n+s^{2n})/n$. If $|s|<1$, its absolute value is at most $3|s{|}^n+|s{|}^{2n}$, and the two geometric series converge. Thus there is absolute convergence.

For $s=1$ the series becomes $\sum 4/n$, a divergent multiple of the harmonic series. For $s=-1$ it becomesThe first series converges by the alternating series test, while the second diverges to $+\infty$, so their partial sums tend to $+\infty$. In particular, alternation of some terms does not imply convergence here. The complete classification isThere are no conditionally convergent cases.

Let $r=|\sin x|\in [0,1]$. For $r<1$, the absolute summand $r^n/\sqrt{n}$ is at most $r^n$, so a geometric series comparison proves absolute convergence. For $r=1$, the series is $\sum (-1{)}^n/\sqrt{n}$: its positive term sizes decrease to zero, so the alternating series test proves convergence. Its absolute series is a divergent p-series with exponent $1/2$. HenceIt converges for every real $x$ and never diverges.

For fixed real $x$, define $u_0=x$ and $u_n=\sin (0.99u_{n-1})$. This makes $u_n$ exactly the summand containing $n$ nested sines. The elementary inequality $|\sin t|\le |t|$ gives inductivelyThe fact that an iterated contraction gives a summable orbit applies here. Comparison with a geometric series therefore givesThus the nested-sine series converges absolutely for every real $x$; there are no divergent or merely conditionally convergent cases.