Keep calm and do the limit

Algebra Level 5

$\large\dfrac{\displaystyle\sum_{k=1}^{n-1}x_kx_{k+1}}{\displaystyle\sum_{k=1}^{n}x_k^2}$

Let $M_n$ be the maximum of the expression above, where $x_1,\ldots,x_n$ are positive real numbers. Find $\displaystyle \lim_{n\to\infty}M_n$ .

Bonus question : Find a closed formula for $M_n$ .

Note : The limit is much easier to find than $M_n$ itself.

The answer is 1.000.

1 solution

Andreas Wendler
Jan 8, 2016

For $x_{1}=x_{2}=...=x_{n}$ the value is $\frac {n-1}{n}$ . But we find greater maxima for example $\frac{\sqrt{2}}{2}$ for n=3 or $\frac{15+\sqrt{5}}{20}$ for n=4. The value of the expression always has to be less than 1 and so does the maximum. Therefore we can write:

$\frac {n-1}{n}<=M_{n}<1$

This means the limit will be equal 1.

Ja genau! (+1) Just to make the solution complete, can you explain why the expression always has to be less than 1.

Where did you get $M_4=\frac{15+\sqrt{5}}{20}$ from? That's not the value I found.

Otto Bretscher - 5 years, 5 months ago

1.) Completion of proof:

First we assume an order: $|x_{1}|<=|x_{2}|<=|x_{3}|<=...<=|x_{n}|$

So we can write: $x_{k}x_{k+1}<=x_{k+1}^{2}$

Therefore: $\sum_{k=1}^{n-1}x_{k}x_{k+1}<=\sum_{k=2}^{n}x_{k}^{2}<\sum_{k=1}^{n}x_{k}^{2}$

q.e.d.

2.) Remark: The mentioned value for $M_4$ I got after calculation by hand using that first derivations to $x_{k}$ must be 0. Maybe that an error occured here.

Andreas Wendler - 5 years, 5 months ago

I don't mean to be difficult, but I don't think we can assume an ascending order of the variables since the function $\sum_{k=1}^{n-1}x_kx_{k+1}$ is not symmetric. At the maximum, the variables will not be ordered the way you assume.

One way to do it is to apply Cauchy-Schwarz to the two vectors $(x_1,x_2,...,x_n,0)$ and $(0,x_1,...,x_{n-1},x_n)$ ; that gives $\sum_{k=1}^{n-1}x_kx_{k+1}<\sum_{k=1}^{n}x_k^2$ so $M_n<1$ .

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – I got the limit the same way ! Can you give any hint on the bonus question?

Utsav Bhardwaj - 5 years, 5 months ago

Keep calm and do the limit

The answer is 1.000.

1 solution

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