My recent work in mathematical analysis has moved from the definition of a limit to properties and concrete calculations involving sequence limits. In the previous stage, I was repeatedly writing epsilon-N proofs. Over the past few days, the main question has instead been how to choose between squeezing, estimation, rationalisation, and a counterexample.
The squeeze theorem is not only a fixed routine
I first reviewed the squeeze theorem: if a sequence lies between two sequences that approach the same limit, then the middle sequence approaches that limit as well.
Several exercises contained factors such as sin n and cos√n, which do not converge but remain bounded. One example was
n^(2/3) sin n / (n+1).
I can use |sin n|≤1 and squeeze the expression. I can also observe that n^(2/3)/(n+1)→0 and use the fact that an infinitesimal multiplied by a bounded quantity is still infinitesimal. I specifically asked whether the squeeze theorem was compulsory here. My current understanding is that the second presentation is shorter, but the ordinary product rule for limits cannot be applied because sin n itself has no limit. Its boundedness is the condition doing the work.
With radicals, first look for a useful rewrite
For √(n+1)-√(n+2), rationalisation turns the difference of two large radicals into
-1/(√(n+1)+√(n+2)),
which makes the convergence to zero much easier to see. For ⁿ√(n²+n), I used n²≤n²+n≤2n², took nth roots, and squeezed the expression to obtain the limit 1.
I also practised the limit of a fixed kth root. If aₙ→a>0, the difference-of-powers identity rewrites the difference between the roots as |aₙ-a| divided by a denominator with a fixed positive lower bound. More precisely, k is fixed and the conclusion is ᵏ√aₙ→ᵏ√a.
Divergent sequences require counterexamples
The most useful part of this topic was working with counterexamples. The sum of two divergent sequences need not diverge: n+(-n)=0. Their product need not diverge either: (-1)ⁿ·(-1)ⁿ=1.
If xₙ converges and yₙ diverges, then xₙ+yₙ must diverge; otherwise subtracting xₙ would force yₙ to converge. The product, however, may still converge, as (1/n)·n=1 shows. Only when xₙ→a≠0 can convergence of the product be used to divide by xₙ, force yₙ to converge, and obtain a contradiction.
These examples showed me that the phrase “one converges and one diverges” is not enough to determine the result of an operation. A zero limit, a nonzero limit, and boundedness lead to different conclusions.
I can now follow the conditions and choose an appropriate method, but I would not call the process automatic yet. My next step is to practise two habits: before calculating, decide whether the target calls for a bound or an equivalent rewrite; and when a statement asks whether something must be true, try the simplest counterexample before attempting a proof.
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