Cfrgtkky

For a monotonically increasing f(x), why does the summation below, on the left hand side, always approximate the lower bound of the summation on the right hand side?

$int_{k=m-1}^n f(k) leq sum_{k=m}^n f(k)$

If $f(x)=k$, $m=2$, and for $ngeq 2$ then the left hand summation is no longer a lower bound, which contradicts the above.

This formula sheet from a class suggests the left hand side would be a lower bound: https://www.cs.umd.edu/class/fall2018/cmsc351-01XX02XX/files/info.pdf

EDIT: I understand that I'm going wrong somewhere but I don't know where. The purpose of my comments is to demonstrate my reasoning for why I think there's a contradiction so that others can point out where my reasoning is incorrect. Thanks

EDIT 2: I've changed the summation symbol to an integral symbol to avoid confusion. It's worth nothing the results would be the same since the functions are called on integers (discrete intervals).

EDIT 3: The result is not the same with the change from a summation to integral because the $n$ on the left hand side is not included in the summation (so same as summation $m-1$ to $n-1$). Thanks Ross!

The upper limit on the left should be $n-1$. The integral you are approximating is from $m-1$ to $n$. You have ends and break points at $m-1,m,m+1,m+2ldots n-1,n$. Using the left side takes the value of $f(x)$ from the left hand end of each interval while the right side takes the value of $f(x)$ from the right hand side of each interval. As $f(x)$ is increasing, the right side will be larger. The true integral will be between the two values.

Your formula sheet gives for $f(x)$ increasing
$$int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)le int_{m}^{n+1}f(x)dx$$
When you try to bound the left integral by a sum you need to make the limits on the integral match. If you shift the limits on the right integral down by $1$ you get the left integral. The lower bound sum then has to have both its limits decreased by $1$, giving the correct formula
$$sum_{k=m-1}^{n-1}f(k) le int_{m-1}^nf(x)dx le sum_{k=m}^nf(k)$$
This shows the upper limit on the left sum should be $n-1$. The easiest way to see the problem is to use $f(x)=1$. Your left sum is then $n-m+1$ and your right sum is $n-m$. The integral is also $n-m$. You are adding one too many terms on the left.