Question: "Given a1=1, an+1=an+1an, find limn→∞ann"

Swebaceacichegh

Answered question

2022-01-21

Given

a_{1} = 1, a_{n + 1} = a_{n} + \frac{1}{a_{n}}

, find

lim_{n \to \infty} \frac{a_{n}}{n}

Jordyn Horne

Beginner2022-01-22Added 16 answers

I completely forgot about the Stolz-Cesàro theorem, from which we get:

lim_{n \to \infty} \frac{a_{n}}{n} = lim_{n \to \infty} \frac{a_{n + 1} - a_{n}}{(n + 1) - n} = lim_{n \to \infty} \frac{\frac{1}{a_{n}}}{1} = lim_{n \to \infty} \frac{1}{a_{n}} = 0

The same technique works for

\frac{a_{n}^{2}}{n}

enguinhispi

Beginner2022-01-23Added 15 answers

a_{n + 1}^{2} = a_{n}^{2} + 2 + \frac{1}{a_{n}^{2}}

, we know that

a_{n + 1}^{2} \leq a_{n}^{2} + 3

a_{n}^{2} \leq 3 n

, then

a_{n + 1}^{2} \leq 3 (n + 1)

, and since

a_{1}^{2} = 1 \leq 3

, by the induction hypothesis

a_{n}^{2} \leq 3 n

Thus,

\frac{a_{n}^{2}}{n^{2}} \leq \frac{3}{n}

for all n, and hence

lim_{n \to \infty} \frac{a_{n}^{2}}{n^{2}} = 0

from this it follows though that

lim_{n \to \infty} \frac{a_{n}}{n} = 0

RizerMix

Expert2022-01-27Added 656 answers

Lets rewrite:

\frac{a_{n + 1} - a_{n}}{n + 1 - n} = \frac{1}{a_{n}}

Now, as n goes to infinity lets denote

a_{n} = f

Then approximately:

\frac{d f}{d n} = \frac{1}{f}

After integrating

\frac{f^{2}}{2} = n + c \Rightarrow \frac{f^{2}}{n^{2}} = \frac{2}{n} + \frac{c}{n^{2}}

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