Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 19 / a

For every integer $n\ge 1$, $\sin ((2n-1)\pi )=0$ and $\cos ((2n-3)\pi )=-1$. Thus the summand simplifies to $2/n$. The resulting series is twice the harmonic series, so it diverges to $+\infty$.

The inner geometric series has common ratio $2/3$, soThe outer geometric series therefore converges as well, with valueAll summands are nonnegative, so there is no cancellation or conditional-convergence issue in evaluating the nested sums.

The summands repeat the six-term patternIn particular they do not tend to zero: the subsequence with $n=6k+1$ is always $\sqrt{3}/2$. The term test for divergence therefore shows that the series diverges. Although each block of six sums to zero, that only makes a subsequence of partial sums zero; it does not make the full sequence of partial sums convergent.

For the positive terms $u_n=2^n/n!$, the consecutive ratio is $u_{n+1}/u_n=2/(n+1)\to 0$, so the ratio test proves convergence. The exponential series gives $e^2={\sum }_{n=0}^{\infty }{2}^n/n!$. Removing the $n=0$ term yields