Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 11D (Analysis I)

The two preliminary sequence limit laws follow directly from the triangle inequality. For the sum, choose indices so that $|x_n-x|<\epsilon /2$ and $|y_n-y|<\epsilon /2$ beyond their respective cutoffs. Beyond the larger cutoff,Thus

For the reciprocal, convergence to nonzero $x$ gives $|x_n-x|<|x|/2$ eventually. The reverse triangle inequality then gives $|x_n|\ge |x|/2$, and consequentlyChoosing the eventual error smaller than both $|x|/2$ and $\epsilon |x{|}^2/2$ provesThe nonzero limit gives the denominator a positive lower bound; nonzero individual terms alone would not do so. The assumption $x_n\ne 0$ ensures the reciprocal sequence is defined even before this eventual bound applies.

Use rationalization of a square-root difference to avoid subtracting two large, nearly equal quantities:Since $1/n\to 0$ and the square root is continuous at $1$, the denominator tends to $2$. The reciprocal limit law established above is applicable because this limit is nonzero. HenceThe limit is one half; rationalization exposes the finite difference hidden by cancellation.

Rationalize the numerator of the summand:For every $n\ge 1$ we have $\sqrt{1+1/n}\le \sqrt{2}$, and thereforeThe harmonic series diverges, so the comparison test for series makes the given positive-term series diverge as well:For completeness, divergence of the harmonic series follows by grouping the terms with $2^k<n\le 2^{k+1}$: each such block contributes at least $2^k/2^{k+1}=1/2$. The same rationalization also shows that $n$ times the summand tends to $1/2$. The summands tend to zero, but their harmonic-size tail prevents convergence.