# What is the value of lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2?

Jan 5, 2017

$1$

#### Explanation:

We have ${\sum}_{k = 1}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 \le {\sum}_{k = 1}^{n} \frac{2 n + 1}{{\left(n + k + 1\right)}^{2} - 1}$ but

$\frac{1}{{\left(n + k + 1\right)}^{2} - 1} = \frac{1}{2} \left(\frac{1}{n + k} - \frac{1}{n + k + 2}\right)$ and

${\sum}_{k = 1}^{n} \frac{1}{{\left(n + k + 1\right)}^{2} - 1} = \frac{1}{2} \left(\frac{1}{n + 1} + \frac{1}{n + 2} - \frac{1}{2 n + 1} - \frac{1}{2 n + 2}\right)$ or more compactly

${\sum}_{k = 1}^{n} \frac{1}{{\left(n + k + 1\right)}^{2} - 1} = \frac{1}{4} \frac{4 {n}^{2} + 5 n}{2 {n}^{3} + 7 {n}^{2} + 7 n + 2}$

now we have

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{2 n + 1}{{\left(n + k + 1\right)}^{2} - 1} = {\lim}_{n \to \infty} \frac{1}{4} \frac{\left(2 n + 1\right) \left(4 {n}^{2} + 5 n\right)}{2 {n}^{3} + 7 {n}^{2} + 7 n + 2} = 1$

so

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 \le 1$

or

${\lim}_{n \to \infty} {\sum}_{k = 0}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 - {\lim}_{n \to \infty} \frac{2 n + 1}{n + 1} ^ 2 \le 1$ so

${\lim}_{n \to \infty} {\sum}_{k = 0}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 \le 1$

Now ${\int}_{\xi = 0}^{\xi = n} \frac{d \xi}{n + \xi + 1} ^ 2 \le {\lim}_{n \to \infty} {\sum}_{k = 0}^{n} \frac{1}{n + k + 1} ^ 2$

but
${\int}_{\xi = 0}^{\xi = n} \frac{d \xi}{n + \xi + 1} ^ 2 = \frac{1}{n + 1} - \frac{1}{2 n + 1}$ and

${\lim}_{n \to \infty} \left(2 n + 1\right) \left(\frac{1}{n + 1} - \frac{1}{2 n + 1}\right) = 1$

so

$1 \le {\lim}_{n \to \infty} {\sum}_{k = 0}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 \le 1$ and concluding

${\lim}_{n \to \infty} {\sum}_{k = 0}^{n} \frac{2 n + 1}{n + k + 1} ^ 2 = 1$