Question

Salut, aveti o idee la problema 279...?Am incercat insa nu-mi iasa...

Answer 1

$\sum\limits_{k=1}^n \dfrac{2k+1}{k^2(k+1)^2} = \sum\limits_{k=1}^n\dfrac{(k+1)^2-k^2}{k^2(k+1)^2}= \sum\limits_{k=1}^n \dfrac{(k+1)^2}{k^2(k+1)^2}-\sum\limits_{k=1}^n \dfrac{k^2}{k^2(k+1)^2}= \\ \\=\sum\limits_{k=1}^n \dfrac{1}{k^2} -\sum\limits_{k=1}^n \dfrac{1}{(k+1)^2} = \\ \\ =\dfrac{1}{1}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}$

$=1-\dfrac{1}{(n+1)^2}\\ \\ \Rightarrow \lim\limits_{n\to \infty}\sum\limits_{k=1}^n \dfrac{2k+1}{k^2(k+1)^2}= \lim\limits_{n\to \infty}\Big(1-\dfrac{1}{(n+1)^2}\Big) = 1-0= \boxed{1}$

Answer 2

Răspuns:

Explicație pas cu pas: