Partea intreaga a unei sume (clasic).
Moderators: Bogdan Posa, Laurian Filip
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Virgil Nicula
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Partea intreaga a unei sume (clasic).
Sa se arate ca \( (\forall )\ n\in\mathbb N^* \) avem \( \left[\sum_{k=1}^n\frac {1}{k^2}\right]\ =\ 1 \) .
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Laurian Filip
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o solutie ar fi folosind metoda inductiei. stiu ca nu este chiar de clasa a 7a, dar nu-mi vine prin minte alta mai simpla.
presupunem
P(k): \( \sum_{k=1}^n\frac {1}{k^2} \leq 2 - \frac{1}{k} \) adevarata
demonstram \( P(k) \to P(k+1) \)
\( \sum_{k=1}^{n+1}\frac {1}{k^2} \leq 2 - \frac{1}{k}+ \frac{1}{{k+1}^2} \leq 2 - \frac{1}{k+1} \) - adevarat.
\( P(1): 1 \leq 1 \) - adevarat.
Cum P(k) implica P(k+1), si P(1) este adevarata, P(n) este adevarata, \( \forall n\in\mathbb{N}^* \)
deci \( 1 \leq \sum_{k=1}^n\frac {1}{k^2} < 2 \)
presupunem
P(k): \( \sum_{k=1}^n\frac {1}{k^2} \leq 2 - \frac{1}{k} \) adevarata
demonstram \( P(k) \to P(k+1) \)
\( \sum_{k=1}^{n+1}\frac {1}{k^2} \leq 2 - \frac{1}{k}+ \frac{1}{{k+1}^2} \leq 2 - \frac{1}{k+1} \) - adevarat.
\( P(1): 1 \leq 1 \) - adevarat.
Cum P(k) implica P(k+1), si P(1) este adevarata, P(n) este adevarata, \( \forall n\in\mathbb{N}^* \)
deci \( 1 \leq \sum_{k=1}^n\frac {1}{k^2} < 2 \)
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Laurian Filip
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fie p astfel incat \( 2^{p}>n \)
\( 1\leq \sum_{k=1}^n\frac {1}{k^2}< \sum_{k=1}^{2^p}\frac {1}{k^2} \)
\( \sum_{k=1}^{2^{p}}\frac {1}{k^2} = 1 + \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2^{p}}^2} < 1 + \frac{1}{2^2} \cdot 2 +\frac{1}{2^4} \cdot 4 +... + \frac{1}{(2^{p-1})^2} \cdot 2^{p-1} = 1 + \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^k} = \frac{1-\frac{1}{2^{p}}}{\frac{1}{2}}=2- \frac{1}{2^{p-1}}
\)
\( 1\leq \sum_{k=1}^n\frac {1}{k^2}< \sum_{k=1}^{2^p}\frac {1}{k^2}\leq2- \frac{1}{2^{p-1}} <2 \)
\( 1\leq \sum_{k=1}^n\frac {1}{k^2}< \sum_{k=1}^{2^p}\frac {1}{k^2} \)
\( \sum_{k=1}^{2^{p}}\frac {1}{k^2} = 1 + \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{2^{p}}^2} < 1 + \frac{1}{2^2} \cdot 2 +\frac{1}{2^4} \cdot 4 +... + \frac{1}{(2^{p-1})^2} \cdot 2^{p-1} = 1 + \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^k} = \frac{1-\frac{1}{2^{p}}}{\frac{1}{2}}=2- \frac{1}{2^{p-1}}
\)
\( 1\leq \sum_{k=1}^n\frac {1}{k^2}< \sum_{k=1}^{2^p}\frac {1}{k^2}\leq2- \frac{1}{2^{p-1}} <2 \)