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\displaystyle\sum_{k=1}^{\infty}\frac{q^k}{k(k+1)}=\sum_{k=1}^{\infty}(\frac{q^k}{k}-\frac{q^k}{k+1})=\sum_{k=1}^{\infty}\frac{q^k}{k}-\frac{1}{q}(\sum_{k=1}^{\infty}\frac{q^{k+1}}{k+1}+q-q)=\sum_{k=1 ...

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Tema: zadaci za vjezbu
holidayin Odgovori: 12 Pogledano: 3845	Forum: Vjerojatnost Postano: 2:41 pon, 19. 1. 2009 Naslov: zadaci za vjezbu
holidayin Odgovori: 12 Pogledano: 3845	\displaystyle\sum_{k=1}^{\infty}\frac{q^k}{k(k+1)}=\sum_{k=1}^{\infty}(\frac{q^k}{k}-\frac{q^k}{k+1})=\sum_{k=1}^{\infty}\frac{q^k}{k}-\frac{1}{q}(\sum_{k=1}^{\infty}\frac{q^{k+1}}{k+1}+q-q)=\sum_{k=1 ...

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