Matematické Fórum

stuart clark · 22. 11. 2011 09:30

floor function sum

vanok · 22. 11. 2011 17:21 — Editoval vanok (22. 11. 2011 18:26)

Hi
Let S the sum to find.
This exercise it resolves very simply thanks to the following inequalities :
$1\le \sqrt[k]{\frac k {k-1}}=\sqrt[k]{1+\frac1{k-1}}\le 1+\frac1{k.(k-1)}=1-\frac1k+\frac1{k-1}$
All the terms eliminate two-two except both extrêmes.
By using them for $k=2,...,2010$ and by adding we obtain:
$2009 \le S \le A \le 2009 +1-\frac1{2010}$

And finally $S=2009$

stuart clark

↑ vanok:

thanks vanok

but how can i prove $1\le \sqrt[k]{\frac k {k-1}}=\sqrt[]{1+\frac1{k-1}}\le 1+\frac1{k.(k-1)}$

vanok · 22. 11. 2011 17:49 — Editoval vanok (22. 11. 2011 18:55)

↑ stuart clark:,
For example by means of the Bernoulli's inequality:

The inequality states that

$1+kx \le (1+x)^k$

for every integer k≥ 0 and every real number x ≥ −1.

For $x=\frac1{k.(k-1)}$ we obtain our result.

Matematické Fórum

#1 22. 11. 2011 09:30

floor function sum

#2 22. 11. 2011 17:21 — Editoval vanok (22. 11. 2011 18:26)

Re: floor function sum

#3 22. 11. 2011 17:35 — Editoval stuart clark (22. 11. 2011 17:40)

Re: floor function sum

#4 22. 11. 2011 17:49 — Editoval vanok (22. 11. 2011 18:55)

Re: floor function sum

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