# Evaluating a limit by Riemann sums

I'm having some issues with understanding exactly how to see that a given limit/sum is a Riemann sum. For example:

$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{k=1}^n \frac{1}{\sqrt{k}}$.

I recognize that $\frac{1}{\sqrt{n}}$ is the length of the partitions, and the other fraction is $f(c_k)$, where $c_k$ is a value picked in the k'th interval of the n'th partition, but to me it looks like $f(x) = \frac{1}{\sqrt{x}}$ and so $c_k$ should be just $k$, but $k$ is a lot bigger than most of the partitions' length.

Another type of problem I see is where $c_k$ depends on $n$, and again it seems to me that there isn't a $c_k$ for each k'th interval. For example this problem:

$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n sin(\frac{k\pi}{2n})$.

Here the length of each partition is $\frac{1}{n}$, but $c_k = \frac{k\pi}{2} \frac{1}{n}$, and since the "multiplier" is greater than one, it looks to me like $c_k$ is bound to skip some intervals eventually, and so there isn't a $c_k$ to each interval, and then it's no longer a Riemann sum.

But, obviously it is, I just don't understand it.. can anyone explain this to me? My textbook has a few problems like this, but no explanations. Thanks a lot in advance.

• I don't follow what you are saying about the skipping of intervals. – copper.hat Oct 16 '14 at 22:44
• As an aside, the function $f$ above is not Riemann integrable on $[0,1]$. It exists as an improper integral, so a little care needs to be taken with the argument. – copper.hat Oct 17 '14 at 1:13

Look at ${1 \over n} \sum_{k=1}^n \sqrt{n \over k} = {1 \over n} \sum_{k=1}^n {1 \over \sqrt{k \over n}} ={1 \over n} \sum_{k=1}^n f({k \over n})$, where $f(x) = {1 \over \sqrt{x}}$.
• Apologies for being dense. But: for each n, we have a partition of [a,b], where the length of each sub-interval of the partition is (in your example) $\frac{1}{n}$. And in each sub-interval we should have a point(in your example, $\frac{k}{n}$), so we can sum the rectangles over each sub-interval. But in the examples I posted, the choices of these points don't seem to fit into these sub-intervals.. if the length of each sub-interval is $\frac{1}{\sqrt{n}}$, then surely if we pick points $1,2,..k$ these points aren't actually in all of these sub-intervals, if $n \geq 2$? – Zak Laberg Oct 17 '14 at 8:47
• Your example is the same as mine. ${1 \over \sqrt{n}} \sum_k {1 \over \sqrt{k}} = {1 \over n} \sum_{k=1}^n \sqrt{n \over k}$. – copper.hat Oct 17 '14 at 14:55
• Ah yes. So the idea seems to be to relate the partition length and the point picked inside it, and then recognize it as a Riemann sum? My second example would then be $\frac{1}{n} \sum sin(\frac{k\pi}{2n}) = \frac{\pi}{2n} \sum \frac{2}{\pi} sin(\frac{k\pi}{2n}) = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} sin(x) dx = \frac{2}{\pi}$. Is that reasoning correct..ish? I'm not sure why it's not Riemann integrable, but your patience has probably run out anyway - thanks a lot for your help! – Zak Laberg Oct 17 '14 at 17:41
• The function $f(x) = {1 \over \sqrt{x}}$ is unbounded on $[0,1]$, so it is not Riemann integrable. However, the limit $\lim_{t\downarrow 0} \int_t^1 f(x) dx$ exists so it has an improper Riemann integral. It means there are a few more details (basically showing that ignored points don't matter) to showing equivalence with the integral. – copper.hat Oct 17 '14 at 17:45