1) Enuntul este cu siguranta incomplet: lipseste 1/(1*2) de la inceput pentru a obtine egalitatea ceruta. Asadar, completand "veriga lipsa", avem:
[tex] \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{2011*2012} [/tex] =
=[tex] \frac{2-1}{1*2} + \frac{3-2}{2*3} + \frac{4-3}{3*4} + ... + \frac{2012-2011}{2011*2012} [/tex] =
=[tex] \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{2011} - \frac{1}{2012} [/tex] =
=[tex] \frac{1}{1} - \frac{1}{2012} [/tex]
=
=[tex] \frac{2011}{2012} [/tex]
2) Avem de aratat ca:
[tex] \frac{1}{10} < \frac{1}{90} + \frac{1}{91} + \frac{1}{92} + ... + \frac{1}{99} < \frac{1}{9} [/tex]
Observam ca avem de aratat ca suma a 10 fractii este cuprinsa intre doua fractii.
Amplificand cu 10 fractiile [tex] \frac{1}{10} [/tex] si [tex] \frac{1}{9} [/tex] obtinem expresia echivalenta:
[tex] \frac{10}{100} < \frac{1}{90} + \frac{1}{91} + \frac{1}{92} + ... + \frac{1}{99} < \frac{10}{90} [/tex]
Avem:
[tex] \frac{1}{100} < \frac{1}{90} < \frac{1}{90} [/tex]
[tex] \frac{1}{100} < \frac{1}{91} < \frac{1}{90} [/tex]
[tex] \frac{1}{100} < \frac{1}{92} < \frac{1}{90} [/tex]
.....................
[tex] \frac{1}{100} < \frac{1}{99} < \frac{1}{90} [/tex]
si daca insumam membru cu membru cele 10 inegalitati de mai sus obtinem:
[tex] \frac{10}{100} < \frac{1}{90} + \frac{1}{91} + \frac{1}{92} + ... + \frac{1}{99} < \frac{10}{90} [/tex]
adica ceea ce am vazut la inceput ca este echivalent cu ceea ce se cerea sa se demonstreze.
3) Avem de aratat ca:
[tex] \frac{1}{ 3^{2} } + \frac{1}{ 4^{2} } + ... + \frac{1}{ 201^{2} } < \frac{1}{2} [/tex]
Cum:
[tex] \frac{1}{ 3^{2} } < \frac{1}{2*3} [/tex] = [tex] \frac{1}{2} - \frac{1}{3} [/tex]
[tex] \frac{1}{ 4^{2} } < \frac{1}{3*4} [/tex] = [tex] \frac{1}{3} - \frac{1}{4} [/tex]
......................
[tex] \frac{1}{ 201^{2} } < \frac{1}{200*201} [/tex] = [tex] \frac{1}{200} - \frac{1}{201} [/tex]
insumand membru cu membru inegalitatile de mai sus obtinem:
[tex] \frac{1}{ 3^{2} } + \frac{1}{ 4^{2} } + ... + \frac{1}{ 201^{2} } < \frac{1}{2} - \frac{1}{201} < \frac{1}{2} [/tex]
(c.c.t.d.)
4) Avem de aratat ca:
[tex] \frac{1}{ 2^{2} } +
\frac{1}{ 4^{2} } +
\frac{1}{ 6^{2} } + ... + \frac{1}{ 2006^{2} } < \frac{2005}{4012} [/tex]
Dam factor comun pe [tex] \frac{1}{ 2^{2} } [/tex] in ambii membri si obtinem:
[tex] \frac{1}{ 2^{2} } *( \frac{1}{ 1^{2} } + \frac{1}{ 2^{2} } +
\frac{1}{ 3^{2} } + ... + \frac{1}{ 1003^{2} } )< \frac{1}{ 2^{2} } * \frac{2005}{1003} [/tex]
si impartind ambii membri la [tex] \frac{1}{ 2^{2} } [/tex] obtinem relatia echivalenta (pe care o avem de demonstrat):
[tex] \frac{1}{ 1^{2} } + \frac{1}{ 2^{2} } +
\frac{1}{ 3^{2} } + ... + \frac{1}{ 1003^{2} } < \frac{2005}{1003} [/tex]
Avem:
[tex] \frac{1}{ 2^{2} } < \frac{1}{1*2} [/tex] = [tex] \frac{1}{1} - \frac{1}{2} [/tex]
[tex] \frac{1}{ 3^{2} } < \frac{1}{2*3} [/tex] = [tex] \frac{1}{2} - \frac{1}{3} [/tex]
[tex] \frac{1}{ 4^{2} } < \frac{1}{3*4} [/tex] = [tex] \frac{1}{3} - \frac{1}{4} [/tex]
.....................
[tex] \frac{1}{ 1003^{2} } < \frac{1}{1002*1003} [/tex] = [tex] \frac{1}{1002} - \frac{1}{1003} [/tex]
si insumand inegalitatile de mai sus, impreuna cu [tex] \frac{1}{ 1^{2} } [/tex] obtinem:
[tex] \frac{1}{ 1^{2} } + \frac{1}{ 2^{2} } +
\frac{1}{ 3^{2} } + ... + \frac{1}{ 1003^{2} } [/tex] < [tex] \frac{1}{ 1^{2} } + \frac{1}{1} - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + ... + \frac{1}{1002} - \frac{1}{1003} [/tex]
adica:
[tex] \frac{1}{ 1^{2} } + \frac{1}{ 2^{2} } +
\frac{1}{ 3^{2} } + ... + \frac{1}{ 1003^{2} } [/tex] < [tex] \frac{1}{1} + \frac{1}{1} - \frac{1}{1003} [/tex] = [tex]2 - \frac{1}{1003} [/tex] =[tex] \frac{2005}{1003} [/tex]
(c.c.t.d.)
