 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-ii/2/b/solution

Put

u_{n} = 1 1^{n} lo g n / (2^{2 n + 1} n!)

for

n \geq 2

. The factorial gives

\frac{u _{n + 1}}{u _{n}} = \frac{11}{4 ( n + 1 )} \frac{l o g ( n + 1 )}{l o g n} .

The logarithm identity

\frac{l o g ( n + 1 )}{l o g n} = 1 + \frac{l o g ( 1 + 1/ n )}{l o g n} ⟶ 1

follows from continuity of the logarithm at one and

lo g n \to + \infty

. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some

N

satisfies

0 < u_{n + 1} \leq u_{n} /2

for all

n \geq N

0 < u_{N + j} \leq u_{N} 2^{- j}

(0, 1)

tends to zero, so the squeeze theorem proves

n \to \infty lim \frac{1 1 ^{n} lo g n}{2 ^{2 n + 1} n !} = 0.

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