Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 19

For $r\ne 1$, differentiate the finite geometric series and multiply by $r$:The required terms are $-n{r}^n$ with $r=-1/3$. The finite weighted geometric sum therefore givesSince $N/3^N\to 0$, the correction tends to zero, proving convergence and yieldingIn fact the ratio test also proves absolute convergence, since the ratio of consecutive absolute terms tends to $1/3$.

By De Moivre's theorem, $(e^{i\alpha }{)}^j=e^{ij\alpha }$. The complex sum whose real part is wanted is therefore a finite geometric series:Use $1-e^{i\beta }=-2i{e}^{i\beta /2}\sin (\beta /2)$ in the numerator and denominator. This yieldsTaking its real part proves the finite trigonometric sumWhen $\alpha \in 2\pi \mathbb{Z}$, the printed quotient is undefined; the original sum is directly $(n+1)\cos \theta$, also the limiting value of the displayed identity. The denominator restriction cannot be omitted from a literal use of the quotient.

Let $a_n=\ln [1+(-1{)}^n/n]$ and $S_N={\sum }_{n=2}^N{a}_n$. Consecutive even and odd terms cancel exactly:ThusBoth parity subsequences have the same limit, so the full sequence of partial sums converges and the sum is zero. This check is necessary: cancellation in fixed pairs alone does not always prove convergence of the original series.

For absolute convergence, each pair instead contributesThe inequality follows from $\ln (1+u)={\int }_0^u(1+s{)}^{-1}ds\ge u/(1+u)$. The lower-bound series diverges by comparison with the harmonic series. Therefore the paired logarithmic series with zero sum has