Question

Fie A = $\frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + ... + \frac{1}{3000}$. Demonstrați că 1 < A < 1,(3).

Answer 1

Răspuns:

1 < A < 1,(3)

Explicație pas cu pas:

împărțim suma în 8 grupe de câte 250 termeni:

$\underbrace{\dfrac{1}{1250} + \dfrac{1}{1250} + ... + \dfrac{1}{1250}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{1001} + \dfrac{1}{1002} + ... + \dfrac{1}{1250}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{1000} + \dfrac{1}{1000} + ... + \dfrac{1}{100}}_{250 \ \ termeni}$

$\iff \dfrac{1}{5} = \dfrac{250}{1250} < S_{1} < \dfrac{250}{1000} = \dfrac{1}{4}$

$\underbrace{\dfrac{1}{1500} + \dfrac{1}{1500} + ... + \dfrac{1}{1500}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{1251} + \dfrac{1}{1252} + ... + \dfrac{1}{1500}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{1250} + \dfrac{1}{1250} + ... + \dfrac{1}{1250}}_{250 \ \ termeni}$

$\iff \dfrac{1}{6} = \dfrac{250}{1500} < S_{2} < \dfrac{250}{1250} = \dfrac{1}{5}$

...

$\underbrace{\dfrac{1}{2750} + \dfrac{1}{2750} + ... + \dfrac{1}{2750}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{2501} + \dfrac{1}{2502} + ... + \dfrac{1}{2750}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{2500} + \dfrac{1}{2500} + ... + \dfrac{1}{2500}}_{250 \ \ termeni}$

$\iff \dfrac{1}{11} = \dfrac{250}{2750} < S_{7} < \dfrac{250}{2500} = \dfrac{1}{10}$

$\underbrace{\dfrac{1}{3000} + \dfrac{1}{3000} + ... + \dfrac{1}{3000}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{2751} + \dfrac{1}{2752} + ... + \dfrac{1}{3000}}_{250 \ \ termeni} < \underbrace{\dfrac{1}{2750} + \dfrac{1}{2750} + ... + \dfrac{1}{2750}}_{250 \ \ termeni}$

$\iff \dfrac{1}{12} = \dfrac{250}{3000} < S_{8} < \dfrac{250}{2750} = \dfrac{1}{11}$

adunăm sumele și obținem:

$\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{10} + \dfrac{1}{11} + \dfrac{1}{12} = \dfrac{28271}{27720} > 1$

și:

$\dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{10} + \dfrac{1}{11} = \dfrac{32891}{27720} < \dfrac{4}{3} = 1.(3)$

de unde rezultă:

$1 < S_{1} + S_{2} + S_{3} + S_{4} + S_{5} + S_{6} + S_{7} + S_{8} < 1,(3)$

$\iff \bf 1 < A < 1,(3)$

q.e.d.