Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 20

Let $a_n=2^n/n!$. The ratio test gives $a_{n+1}/a_n=2/(n+1)\to 0<1$, so the series converges absolutely. In fact the exponential series identifies its value aswith the $n=0$ term subtracted.

Here $0<\ln 2<1$, so $(\ln 2{)}^{-n}$ grows exponentially rather than decaying. The positive terms $a_n=n^3/(\ln 2{)}^n$ do not tend to zero; indeed they tend to $+\infty$. Therefore the series diverges by the necessary condition that the terms of a convergent series tend to zero.

The ratio test gives the same conclusion: $a_{n+1}/a_n=((n+1)/n{)}^3/\ln 2\to 1/\ln 2>1$.

For absolute convergence, examine ${\sum }_{n\ge 1}|x-3{|}^n/n$. When $0<|x-3|<1$, the ratio of successive terms tends to $|x-3|<1$, so the ratio test proves convergence; at $x=3$ every term is zero. For $|x-3|>1$, the terms do not tend to zero.

At either boundary $|x-3|=1$, the absolute-value series is the divergent harmonic series. Thus for real $x$ the complete answer isThe original series at $x=2$ is conditionally convergent by the alternating-series criterion, but it is not absolutely convergent and is correctly excluded here.

The cosine-difference identity gives $\cos x-\cos 3x=2\sin 2x\sin x$. HenceEach sine ratio tends to one, soEquivalently, the quadratic terms of the two cosine expansions differ by $4x^2$.

Near zero, $\cos x=1-x^2/2+O(x^4)$ and $\ln (1+v)=v+O(v^2)$, so $\ln \cos x=-x^2/2+O(x^4)$. Replacing $x$ by $3x$ similarly gives $\ln \cos 3x=-9x^2/2+O(x^4)$. ThereforeFor all sufficiently small nonzero $x$, both cosines are positive and the denominator logarithm is nonzero, so this quotient and its two-sided real limit are well-defined.