Find the limit using Riemann sum

[devinaxxx]

Homework Statement

i want to find limit value using riemann sum
$\lim_{n\to\infty}\sum_{i = 1}^{2n} f(a+\frac{(b-a)k}{n})\cdot\frac{(b-a)}{n}= \int_a^b f(x)dx$<br>
question : <br>
$\lim_{h \to \infty} =\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{2n+(2n-1)}$<br>

Homework Equations

The Attempt at a Solution

<br>
$\lim_{h \to \infty}\sum_{k=1}^n \frac{1}{n}\frac{1}{2+(2k-1)\frac{1}{n}}$
i try to isolate 1/n but i can't find way to make this become $f(\frac{k}{n})$ since k is stuck in $2k-1$, can someone give me a hint? thanks

 

[haruspex]

You can do Riemann sums with unequal widths. See e.g. https://www.muncysd.org/cms/lib/PA06000076/Centricity/Domain/104/LarCalc10_ch04_sec3.ppt

 

[Ray Vickson]

Homework Statement

i want to find limit value using riemann sum
$\lim_{n\to\infty}\sum_{i = 1}^{2n} f(a+\frac{(b-a)k}{n})\cdot\frac{(b-a)}{n}= \int_a^b f(x)dx$<br>
question : <br>
$\lim_{h \to \infty} =\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{2n+(2n-1)}$<br>

Homework Equations

The Attempt at a Solution

<br>
$\lim_{h \to \infty}\sum_{k=1}^n \frac{1}{n}\frac{1}{2+(2k-1)\frac{1}{n}}$
i try to isolate 1/n but i can't find way to make this become $f(\frac{k}{n})$ since k is stuck in $2k-1$, can someone give me a hint? thanks

How different are
$$t_1(n,k) = \frac{1}{2+\frac{2k-1}{n}} \;\; \text{and} \;\; t_2(n,k) = \frac{1}{2 + \frac{2k}{n}} ? $$
Do $\sum \frac{1}{n} t_1(n,k)$ and $\sum \frac{1}{n} t_2(n,k)$ have the same large-$n$ limits?

 

[devinaxxx]

How different are
$$t_1(n,k) = \frac{1}{2+\frac{2k-1}{n}} \;\; \text{and} \;\; t_2(n,k) = \frac{1}{2 + \frac{2k}{n}} ? $$
Do $\sum \frac{1}{n} t_1(n,k)$ and $\sum \frac{1}{n} t_2(n,k)$ have the same large-$n$ limits?

thankou got it!

 

[devinaxxx]

You can do Riemann sums with unequal widths. See e.g. https://www.muncysd.org/cms/lib/PA06000076/Centricity/Domain/104/LarCalc10_ch04_sec3.ppt

thankyou!