Question

O limita la care cer o rezolvare​

Answer 1

$\displaystyle \lim\limits_{n\to \infty}\dfrac{1}{n^{p+1}}\sum\limits_{i=1}^n\dfrac{(p+i)!}{i!}\\ \\ = \lim\limits_{n\to \infty}\dfrac{1}{n^{p+1}}\sum\limits_{i=1}^n\dfrac{i!\cdot (i+1)\cdot (i+2)\cdot ...\cdot (i+p)}{i!}\\ \\ = \lim\limits_{n\to \infty}\dfrac{1}{n^{p+1}}\sum\limits_{i=1}^n\left[(i+1)\cdot(i+2)\cdot ...\cdot (i+p)\right]\\ \\=\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{i=1}^n\dfrac{(i+1)\cdot(i+2)\cdot ...\cdot (i+p)}{n^p} \\\\ =\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{i=1}^n\dfrac{(i+1)\cdot(i+2)\cdot ...\cdot (i+p)}{n\cdot n\cdot \underset{\text{de p ori}}{\underbrace{...}}\cdot n}\\ \\= \lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{i=1}^n\left[\left(\frac{i}{n}+\frac{1}{n}\right)\cdot\left(\frac{i}{n}+\frac{2}{n}\right)\cdot...\cdot \left(\frac{i}{n}+\frac{p}{n}\right)\right]\\ \\ = \lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{i=1}^n\left[\left(\frac{i}{n}+0\right)\cdot\left(\frac{i}{n}+0\right)\cdot\underset{\text{de p ori}}{\underbrace{...}}\cdot \left(\frac{i}{n}+0\right)\right]\\ \\= \lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{i=1}^n\left(\frac{i}{n}\right)^p = \int_{0}^1 x^p\, dx = \dfrac{x^{p+1}}{p+1}\Bigg|_{0}^1 =\dfrac{1^{p+1}}{p+1}-\dfrac{0^{p+1}}{p+1}=\\ \\ = \boxed{\boxed{\dfrac{1}{p+1}}}$