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

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

Let $S_N={\sum }_{n=1}^N(-1{)}^{n+1}/n$. Integrate the finite geometric series identity ${\sum }_{k=0}^{N-1}(-t{)}^k=[1-(-t{)}^N]/(1+t)$ over $0\le t\le 1$:The remainder has absolute value at most ${\int }_0^1{t}^{N}dt=1/(N+1)\to 0$. Hence the partial sums converge, withThis proves both convergence and the value of the alternating harmonic series without substituting an endpoint into an unproved infinite power series. Its convergence is conditional because the corresponding harmonic series of absolute values diverges.