# Riemann sum to definite integral

I'm asked to give a function $f$ such that the sum $$\sum_{i=1}^n \frac{1}{n(2+\frac{i}{n})\ln(2+\frac{i}{n})}$$

is a Riemann-sum on the interval $[0, 1]$. I am then asked to find the limit

$\lim_{n\to \infty}$ $\sum_{i=1}^n \frac{1}{n(2+\frac{i}{n})\ln(2+\frac{i}{n})}$

This is my first year calculus course, and I'm having a hard time understanding how to go from Riemann-sum to definite integral and vice versa. From the general formula $\sum_{i=1}^n f(a+i\Delta x)\Delta x$, I know that $\Delta x = \frac{b-a}{n}$. Now since I am asked to give a function $f$ such that the sum is a Riemann-sum on the interval $[0, 1]$ I believe my $\Delta x = \frac{1}{n}$. Is this correct? If so $i\Delta x$ should yield $i\frac{1}{n} = \frac{i}{n}$, which is present in the sum. I am unsure about where to go from here and get to the function which I can later integrate from 0 to 1...

Any help would be greatly appreciated!

• I doubt the first factor in the denominator is $2$. – Bernard Sep 23 '17 at 13:12
• No, it's n, i've got no idea how it changed to 2. Fixed now – novo Sep 23 '17 at 13:15

It's better to assume $x_i=2+\dfrac{i}{n}$ then $$\Delta x=x_{i+1}-x_i=\dfrac1n$$ and as $n\to\infty$, $x_1=2+\dfrac{1}{n}\to2$ and $x_n=2+\dfrac{n}{n}\to3$ so $$\sum_{i=1}^n \frac{1}{n(2+\dfrac{i}{n})\ln(2+\dfrac{i}{n})}=\color{blue}{\int_2^3\dfrac{dx}{x\ln x}}$$
• Just plug $x_i$! – Nosrati Sep 23 '17 at 13:31
Here's a trick: instead of using the explicit formula for Riemann sums, set $x_i=\dfrac in$. Your sum becomes
$$\sum_{i=1}^n \frac{1}{n(2+x_i)\ln(2+x_i)}=\sum_{i=1}^n \frac{1}{(2+x_i)\ln(2+x_i)}\Delta x,$$ so the function is defined by $\; f(x)=\dotsm$