Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 12

Using the standard sine limit $\sin t/t\to 1$, with angles in radians,Thus the two-sided limit exists and is finite.

Although $\cos (1/x)$ oscillates without a limit, its magnitude is at most one. Therefore $|x\cos (1/x)|\le |x|\to 0$. The squeeze theorem givesThe shrinking factor $x$ suppresses the oscillation from either side of zero.

The derivative of the exponential function at zero gives $e^{-x}=1-x+O(x^2)$. Hence $1-e^{-x}=x+O(x^2)$, andThe denominator is nonzero for every real $x\ne 0$, so this is a two-sided finite limit.

A series ${\sum }_{n=1}^{\infty }{u}_n$ is a convergent series if its partial sums $S_N={\sum }_{n=1}^N{u}_n$ have a finite limit $S$ as $N\to \infty$. The number $S$ is the sum of the series. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The condition $u_n\to 0$ alone is insufficient: the harmonic series has terms tending to zero but diverges.

The series $\sum u_n$ has absolute convergence if $\sum |u_n|$ is a convergent series, meaning its nonnegative partial sums have a finite limit. Absolute convergence implies convergence of the original real or complex series. The converse fails: the alternating harmonic series converges, while its absolute-value series is the divergent harmonic series.

The summands $u_n=n^4/2^n$ are positive. Apply the ratio test:ThereforeFor a direct comparison interpretation, the ratio is eventually at most any fixed number between $1/2$ and $1$, such as $3/4$. The tail is then bounded by a convergent geometric series. Thus the exponential denominator wins over the fourth-power numerator strongly enough to make the whole series summable.