Matematické Fórum

stuart clark · 10. 12. 2011 18:51

Let $S_{n}=\sum_{r=1}^{n}\left(\frac{n^2+nr+r^2}{n^3}\right)$ and

$T_{n}=\sum_{r=0}^{n-1}\left(\frac{n^2+nr+r^2}{n^3}\right)$ for $n=1,2,3,......$

Then Which one is Right

(a) $T_{n}<\frac{11}{6}$

(b) $T_{n}>\frac{11}{6}$

(c) $S_{n}<\frac{11}{6}$

(d) $S_{n}>\frac{11}{6}$

vanok · 10. 12. 2011 23:21 — Editoval vanok (12. 12. 2011 22:04)

Hi ↑ stuart clark:,
To resolve this problem, we shall use two known formulae
$\sum_{r=1}^nr = \frac{n(n+1)}2=\frac{n^2}2 + \frac n 2$
$\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$

$S_{n}=\sum_{r=1}^{n}\left(\frac{n^2+nr+r^2}{n^3}\right)=$
$\sum_{r=1}^{n} \frac {(n+r)^2 - nr}{n^3}=$
$\frac{\sum_{r=1}^{2n}r^2-\sum_{r=1}^{n}r^2- n\sum_{r=1}^{n}r}{n^3}=$
$$\frac{(2n)^3}{3} + \frac{(2n)^2}{2} + \frac{(2n)}{6}-\(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} $ -n$\frac{n^2}2 + \frac n 2$\)\frac 1{n^3}=$
$$\(\frac 83 -\frac 13 - \frac 12$n^3 + $\frac 42 -\frac 12 - \frac 12 $n^2 +$\frac 26 -\frac 16$n\)\frac 1{n^3}=$
$\frac {11}6+\frac 1n +\frac 1{6n^2}$
So we have (d)

Let us notice that
$T_{n}=\sum_{r=0}^{n-1}\left(\frac{n^2+nr+r^2}{n^3}\right)=\frac 1n +S_n - \frac {3n^2}{n^3}=$
$\frac 1n + \frac {11}6 +\frac 1n +\frac 1{6n^2}-\frac 3n=\frac {11}6 -\frac 1n +\frac 1{6n^2}$
We notice that $-\frac 1n +\frac 1{6n^2}=\frac{-6n+1}{6n^2}$ is $<0$ for $n>\frac 16$
So we have (a)

Sincerely Vanok

stuart clark · 11. 12. 2011 10:06

thank vanok

Now I have also a same question, which is given here

Let $S_{n}=\sum_{r=1}^{n}\left(\frac{n}{n^2+rn+r^2}\right)$ and

$T_{n}=\sum_{r=0}^{n-1}\left(\frac{n}{n^2+rn+r^2}\right)$ for $n=1,2,3,......$

Then Which one is Right

(a) $T_{n}<\frac{\pi}{3\sqrt{3}}$

(b) $T_{n}>\frac{\pi}{3\sqrt{3}}$

(c) $S_{n}<\frac{\pi}{3\sqrt{3}}$

(d) $S_{n}>\frac{\pi}{3\sqrt{3}}$

vanok · 12. 12. 2011 12:55

Hi ↑ stuart clark:,
We notice that the expression $S_n$ is a sum of Riemann of the function
$f:f(x)=\frac 1{1 +x +x^2}$ and
$\int_{0}^{1}f(x)dx=\frac{\pi}{3\sqrt{3}}$
(doubtless, it will be necessary to study the monotony of two given sequences).
I shall look at it later.

vanok · 12. 12. 2011 22:34 — Editoval vanok (13. 12. 2011 09:12)

Here is the promised solution.

The sum of Riemann
$S_{n}=\sum_{r=1}^{n}\left(\frac{n}{n^2+rn+r^2}\right)= \frac 1n \sum_{r=1}^{n}\left(\frac{n^2}{n^2+rn+r^2}\right)= \frac 1n \sum_{r=1}^{n}\left(\frac 1{1+\frac rn+(\frac rn)^2}\right)$
corresponds to the choice of right border of each of n equal intervals of length $\frac 1n$ .

And the sum of Riemann
$T_{n}=\sum_{r=0}^{n-1}\left(\frac{n}{n^2+rn+r^2}\right)= \frac 1n \sum_{r=0}^{n-1}\left(\frac{n^2}{n^2+rn+r^2}\right)= \frac 1n \sum_{r=1}^{n}\left(\frac 1{1+\frac rn+(\frac rn)^2}\right)$
corresponds to the choice of left border of each of n equal intervals of length $\frac 1n$ .

As the function f is decreasing on the interval $[0; 1]$ we obtain:
$S_{n}<\int_{0}^{1}f(x)dx=\frac{\pi}{3\sqrt{3}}<T_n$ .
So (b) and (c) are true.