Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 19 / a

For every integer $n$, $\cos ((2n-1)\pi )=-1$. The series is therefore the negative of the harmonic series:For a direct divergence argument, each block $2^{j-1}<n\le 2^j$ contributes at least $2^{j-1}/2^j=1/2$ to the positive sum. Its partial sums grow without bound, soDespite the odd multiples of $\pi$, the signs do not alternate.

For each fixed $n$, the inner geometric series sums to $2^{-n}/(1-1/2)=2^{1-n}$. Summing this second geometric series givesAll terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the double series converges. Equivalently, collecting equal values of $m$ gives ${\sum }_{m=0}^{\infty }(m+1)2^{-m}=4$.