Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 9D (Analysis I)

For a nonzero $x$, the absolute successive-term ratio in the exponential series is $|x|/(n+1)\to 0$, so the first series converges for every $x$. For the factorial series the ratio is $(n+1)|x|\to \infty$, so its terms fail to tend to zero for every $x\ne 0$.

The third series is sparse: its exponent is $n^2$, as in the PDF. If $t_n=(n!{)}^2|x{|}^{n^2}$, thenFor $|x|>1$, the terms grow without bound; at $|x|=1$ their moduli are $(n!{)}^2$, so they again fail the term test. Equivalently, the Cauchy-Hadamard theorem for the sparse coefficients uses $(n!{)}^{2/n^2}\to 1$, not $(n!{)}^{2/n}\to \infty$. Thus the three radii of convergence areZeros at non-square coefficient indices do not change the relevant limsup in the last root test. This illustrates the radius of convergence of a sparse power series.

The Taylor theorem with Lagrange remainder says that, if $f$ has $N+1$ continuous derivatives on the interval joining $0$ and $x$, then for some $\xi$ strictly between them,More generally the expansion is about any base point, replacing $x$ by its displacement. The displayed regularity is a sufficient hypothesis for the theorem.

For $f(t)=(1+t{)}^{1/2}$ and $t>-1$, repeated differentiation givesTherefore the requested coefficients are $c_n=f^{(n)}(0)/n!$. For a fixed $0<x<1$, the remainder hasNow $|c_1|=1/2$ and $|c_{n+1}/c_n|=(n-1/2)/(n+1)<1$ for $n\ge 1$, so $|c_{N+1}|\le 1/2$. Hence $|R_N(x)|\le x^{N+1}/2\to 0$. Passing to the limit proves, rather than merely formally suggests, the binomial seriesThe restriction $x>0$ makes the remainder bound especially direct, since $1+\xi \ge 1$.