encontrar el Límite de $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$

Question

muestran la secuencia de $a_n= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}$ converge a log2

mi intento:

1. secuencia $a_n$ es monótona creciente
2. 0<$a_n$<1/2

cómo encontrar el límite?

Answer 1

$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 {n+i}=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac 1 n\frac 1 {1+\dfrac in}=\int_0^1\frac 1 {1+x}dx=\ln(1+1)-\ln(1+0)=\ln(2)$$ (Suma de Riemann)

Answer 2

$$ \begin{align} \log(2) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\ &=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{(-1)^{n-1}}k\\ &=\lim_{n\to\infty}\left(\vphantom{\sum_{k=1}^n}\right.\overbrace{\sum_{k=1}^{2n}\frac1k}^{\substack{\text{sum of all}\\\text{terms}}}-2\overbrace{\sum_{k=1}^n\frac1{2k}}^{\substack{\text{sum of even}\\\text{terms}}}\left.\vphantom{\sum_{k=1}^n}\right)\\[6pt] &=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\[6pt] &=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\ \end{align} $$

Answer 3

Aquí es clásica prueba de ello. Deje $H_n=\sum_{i=1}^n 1/i$. Es bien conocido que $$ H_n = \ln n + \gamma + \epsilon_n, $$ donde $\gamma$ es de Euler–Mascheroni constante, $\epsilon_n\to 0$. Así, $$ a_n = H_{2n}-H_{n} = \ln 2 + \epsilon_{2n} - \epsilon_n \a \ln2. $$

Answer 4

Aviso, $$a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots +\frac{1}{n+n}$$ $$=\sum_{r=1}^{\infty}\frac{1}{n+r}=\sum_{r=1}^{\infty}\frac{1}{n\left(1+\frac{r}{n}\right)}$$$$=\sum_{i=1}^{\infty}\frac{\frac{1}{n}}{1+\frac{r}{n}}$$ Ahora, vamos a $\frac{r}{n}=x\implies \frac{1}{n}=dx\to 0$ $$\text{upper limit of x}=\lim_{n\to \infty}\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty}\frac{1}{n}=0$$ Hence, using integration we get $$a_n=\int_{0}^{1}\frac{dx}{1+x}$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$ $$=\left[\ln 2-\ln 1\right]=\ln 2$$