Find limit $\lim_{h \to \infty} \frac{1}{2n+1}+\frac{1}{2n+3}+....+\frac{1}{2n+(2n-1)}$ using Riemann sum.

Question

I want to find limit value using Riemann sum $$\lim_{n\to\infty}\sum_{i = 1}^{2n} f(a+\frac{(b-a)k}{n})\cdot\frac{(b-a)}{n}= \int_a^b f(x)dx$$ Question:

$$\lim_{n \to \infty} \frac{1}{2n+1}+\frac{1}{2n+3}+....+\frac{1}{2n+(2n-1)}$$

Attempt:

$$\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{n}\frac{1}{2+(2k-1)\frac{1}{n}}$$ i try to isolate 1/n but i cant find way to make this become $f(\frac{k}{n})$ since k is stuck in $2k-1$, can someone give me a hint? thanks

Answer 1

Hint:)

Let $x_k=\dfrac{2k-1}{n}$ then $\Delta x=\dfrac{2}{n}$.

Edit:(

With $\Delta x=x_{k+1}-x_k=\dfrac{2k+1}{n}-\dfrac{2k-1}{n}=\dfrac{2}{n}$ we have by your notation \begin{align} \lim_{n \to \infty}\sum_{k=1}^n \frac{1}{n}\frac{1}{2+(2k-1)\frac{1}{n}} &= \frac{1}{2}\lim_{n \to \infty}\sum_{k=1}^n \frac{2}{n}\frac{1}{2+\frac{2k-1}{n}}\\ &= \frac{1}{2}\lim_{n \to \infty}\sum_{k=1}^n \frac{1}{2+x_k}\,\Delta x\\ &= \frac{1}{2}\int_0^2\dfrac{1}{2+x}\,dx\\ &= \color{blue}{\ln\sqrt{2}} \end{align}

Answer 2

The above limit can be written as $$\lim_{n \to \infty} 1/n \sum_{k=1}^n \frac{1}{1+2k/n}$$ because when in the limit $n \to \infty$ $1/n=0$.Now standard formula can be applied.