Безкрайна сума

[nikola.topalov]

Да се пресметне сумата $$\sum_{i=1}^{\infty}\ln\left(\dfrac{i(i+2)}{(i+1)^2}\right)$$

[peyo]

Да се пресметне сумата $$\sum_{i=1}^{\infty}\ln\left(\dfrac{i(i+2)}{(i+1)^2}\right)$$

$k = i+1$

$\sum_{i=1}^{\infty}\ln\left(\dfrac{i(i+2)}{(i+1)^2}\right) = \sum_{k=2}^{\infty}\ln\left(\dfrac{(k-1)(k+1)}{k^2}\right) = ln \prod_{k=2}^{\infty} \left(\dfrac{(k-1)(k+1)}{k^2}\right) = ...$

$= ln \dfrac{1.3}{2^2} \dfrac{2.4}{3^2} \dfrac{3.5}{4^2} \dfrac{4.6}{5^2}... = ln \dfrac{1. \cancel 3}{2 . \cancel 2} \dfrac{ \cancel 2. \cancel 4}{ \cancel 3^2} \dfrac{ \cancel 3. \cancel 5}{ \cancel 4^2} \dfrac{ \cancel 4. \cancel 6}{ \cancel 5^2}... = ln \dfrac{1}{2} = -0.6931471805599453$

Да проверим:

In [29]: sum ( math.log( (i*(i+2))/((i+1)**2)) for i in range(1,10000000))
Out[29]: -0.6931470805595877