Examples from Sequences and limits

Proving $2+\frac1{n+1}\to2$

Let $a_n=2+\frac1{n+1}$. Given $\varepsilon>0$, choose an integer $N$ with $N>1/\varepsilon$. If $n>N$, then

Rationalising a square root

For

multiply by the conjugate:

Logarithmic, root and geometric limits

For every $\alpha>0$,

L'Hôpital's rule applied to $\ln x/x^\alpha$ gives

For fixed $x>0$,

For $0<x<1$, geometric decay gives $x^n\to0$. Finally,

Taking logarithms,

and exponentiating gives the result.

A factorial sequence

For

at least $\lfloor n/2\rfloor$ factors are at most $1/2$. Therefore

The rough hierarchy

for fixed $p>0$ and $c>1$ helps choose comparisons, but it is not itself a proof of a limit.

Two monotone sequences

The sequence

is increasing and bounded above by $1$, so it converges. Its limit is $1$.

For

we have

once $n\ge2$. It is eventually decreasing and bounded below by $0$, so it converges. Its limit is $0$, because the ratio is at most $1/2$ for $n\ge3$, so $b_n\le b_3(1/2)^{n-3}$ for all later terms.

A recursively defined sequence

Let

The positive solution of the fixed-point equation $L^2=1+3L$ is

Since $a_1=4\ge\alpha$ and since

the recursion gives $a_{n+1}=\sqrt{1+3a_n}\ge\alpha$. Induction therefore gives $a_n\ge\alpha$ for every $n$. For $a_n\ge\alpha$,

and the latter holds because $a_n^2-3a_n-1\ge0$ for $a_n\ge\alpha$. Thus $\{a_n\}$ is decreasing and bounded below by $\alpha$, so it converges.

If $a_n\to L$, continuity of the square-root function gives

The sequence is positive, so the compatible root is