32E (Asymptotic Methods)

lo g (cos u) \sim - \frac{u ^{2}}{2} (u \to 0),

one has

\frac{ϕ _{n + 1} ( x )}{ϕ _{n} ( x )} \sim x^{2} ⟶ 0.

ii

This is not an asymptotic sequence, because

\frac{ψ _{n + 1} ( x )}{ψ _{n} ( x )} = (\frac{n + 1}{n})^{1/ x} ⟶ 1

x \to \infty

, rather than zero.

sin (x^{n})

oscillate and have arbitrarily large zeros; the ratio

sin (x^{n + 1}) / sin (x^{n})

has no limit, let alone limit zero.

Recall that positivity and asymptoticity mean

\frac{ϕ _{j + 1}}{ϕ _{j}} \to 0, \frac{ψ _{j + 1}}{ψ _{j}} \to 0

for each fixed

j

. For

0 \leq k \leq n

, compare the

k

th term of

χ_{n + 1}

with the

k

th term of

χ_{n}

\frac{ϕ _{k} ψ _{n + 1 - k}}{ϕ _{k} ψ _{n - k}} = \frac{ψ _{n + 1 - k}}{ψ _{n - k}} \to 0.

For the remaining term, compare with the final term of

χ_{n}

\frac{ϕ _{n + 1} ψ _{0}}{ϕ _{n} ψ _{0}} = \frac{ϕ _{n + 1}}{ϕ _{n}} \to 0.

Since

χ_{n}

is positive and dominates every comparison term, summing these finitely many bounds gives

\frac{χ _{n + 1} ( x )}{χ _{n} ( x )} \to 0 .

Thus $(χ_{n})$ is an asymptotic sequence.

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