# Would you please tell me what's the limit of (V_n-U_n) as n approaches positive infinity ? PS: U_(n+1)= sqrt(U_nV_n) and V_(n+1)=(U_n+V_n)/2

Nov 18, 2016

${\lim}_{n \to \infty} \left({V}_{n} - {U}_{n}\right) = 0$

#### Explanation:

Making ${U}_{n} = \lambda {V}_{n}$ and ${\delta}_{n} = {V}_{n} - {U}_{n}$ we have

${\delta}_{n + 1} = {V}_{n + 1} - {U}_{n + 1} = \left(1 - \lambda\right) {V}_{n + 1} = \frac{\lambda + 1}{2} {V}_{n} - \sqrt{\lambda} {V}_{n}$ or

${V}_{n + 1} / \left({V}_{n}\right) = \frac{1}{2} \frac{\lambda + 1 - 2 \sqrt{\lambda}}{1 - \lambda} = \frac{1}{2} {\left(1 - \sqrt{\lambda}\right)}^{2} / \left(1 - {\left(\sqrt{\lambda}\right)}^{2}\right) = \frac{1}{2} \frac{1 - \sqrt{\lambda}}{1 + \sqrt{\lambda}}$

then ${\delta}_{n + 1} = {V}_{0} {\left(\frac{1}{2} \frac{1 - \sqrt{\lambda}}{1 + \sqrt{\lambda}}\right)}^{n}$ so

${\lim}_{n \to \infty} {\delta}_{n} = 0$ because $\frac{1}{2} \left(\frac{1 - \sqrt{\lambda}}{1 + \sqrt{\lambda}}\right) < 1$

Nov 19, 2016

This is my thoughts:

${U}_{n + 1} = \sqrt{{U}_{n} {V}_{n}}$
${V}_{n + 1} = \frac{1}{2} \left({U}_{n} + {V}_{n}\right)$

Let $\delta \left(n\right) = {V}_{n} - {U}_{n}$

Assume the limit exists, and that:
${\lim}_{n \rightarrow \infty} \left({V}_{n} - {U}_{n}\right) = \epsilon$

$\therefore {\lim}_{n \rightarrow \infty} \delta \left(n\right) = \epsilon$

Then it must also be the case that:
$\therefore {\lim}_{n \rightarrow \infty} \delta \left(n + 1\right) = \epsilon$

$\therefore {\lim}_{n \rightarrow \infty} \frac{\delta \left(n + 1\right)}{\delta \left(n\right)} = 1$

Now:
$\delta \left(n + 1\right) = {V}_{n + 1} - {U}_{n + 1}$

$\delta \left(n + 1\right) = \frac{{U}_{n} + {V}_{n}}{2} - \sqrt{{U}_{n} {V}_{n}}$

$\frac{\delta \left(n + 1\right)}{\delta \left(n\right)} = \frac{\frac{{U}_{n} + {V}_{n}}{2} - \sqrt{{U}_{n} {V}_{n}}}{{V}_{n} - {U}_{n}} = 1$

$\frac{{U}_{n} + {V}_{n}}{2} - \sqrt{{U}_{n} {V}_{n}} = {V}_{n} - {U}_{n}$
${U}_{n} + {V}_{n} - 2 \sqrt{{U}_{n} {V}_{n}} = 2 {V}_{n} - 2 {U}_{n}$
$3 {U}_{n} - {V}_{n} = 2 \sqrt{{U}_{n} {V}_{n}}$
${\left(3 {U}_{n} - {V}_{n}\right)}^{2} = {\left(2 \sqrt{{U}_{n} {V}_{n}}\right)}^{2}$
$9 {U}_{n}^{2} - 6 {U}_{n} {V}_{n} + {V}_{n}^{2} = 4 {U}_{n} {V}_{n}$
$9 {U}_{n}^{2} - 10 {U}_{n} {V}_{n} + {V}_{n}^{2} = 0$
$\left(9 {U}_{n} - {V}_{n}\right) \left({U}_{n} - {V}_{n}\right) = 0$

$9 {U}_{n} - {V}_{n} = 0 \implies {U}_{n} = \frac{1}{9} {V}_{n}$
${U}_{n} - {V}_{n} = 0 \implies {U}_{n} = {V}_{n}$

Not sure how this helps, but I welcome comments

Nov 20, 2016

$\lim n \to \infty \left({V}_{n} - {U}_{n}\right)$ is non-negative.

#### Explanation:

${U}_{n + 1} = \sqrt{{U}_{n} {V}_{n}} \to {U}_{n + 1} > 0 \to {U}_{n} > 0 \to {V}_{n} > 0$

Use the property of Means, $G M \le A M$.

Here, $\sqrt{{U}_{n} {V}_{n}} \le \frac{{U}_{n} + {V}_{n}}{2}$

It follows that

${U}_{n + 1} \le {V}_{n + 1}$. So,

${V}_{n} - {U}_{n} \ge 0$, for n=2, 3, 4, ...# So,

$\lim n \to \infty \left({V}_{n} - {U}_{n}\right)$ is non-negative.