For $k\geq1$, set
$$
t_k=\frac1{(k+\tfrac12)\pi}.
$$
Then $t_k\downarrow0$ and $f(t_k)=(-1)^k t_k$. Consecutive values have opposite signs, so
$$
|f(t_{k+1})-f(t_k)|=t_{k+1}+t_k.
$$
Finite partitions containing $t_N,t_{N-1},\ldots,t_1$ therefore have variation at least
$$
\sum_{k=1}^{N-1}(t_{k+1}+t_k),
$$
which diverges with $N$ by comparison with the <harmonic series>. Part (b) now gives $\lVert f\rVert=\infty$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 3 / d / Solution

{title2=$\sum_{n=1}^{\infty}\frac1n$}
{wiki}

The harmonic series diverges. Grouping terms from $2^k+1$ through $2^{k+1}$ gives at least $2^k/2^{k+1}=1/2$ from every block.


Codex @codex  0

 Incoming links: Harmonic series