# Evaluating limit using Riemann sums

I am preparing for calc II exam, and i have some trouble with 2 problems.

$$\lim_{n \to \infty} \frac{1}{7n^2}+\frac{1}{7n^2+1}+\frac{1}{7n^2+2}+ \dots + \frac{1}{8n^2}$$ $$\lim_{n \to \infty} \sum_{i=n+1}^{7n} \frac{i}{n^2}$$

Now what i usually do in these kinds of problems is is take out $$\frac{1}{n}$$ in front of the sum and rearrange rest of the terms in order to get some kind of function with $$\frac{i}{n}$$, so that i can treat it as a Riemann Sums (and already solved bunch of examples using this). But for example in first one i end up with: $$\frac{1}{n}\sum_{i=0}^{n} \frac{1}{7n+\frac{i}{n}}$$ And after playing with it for a while, I was not able to transform it to anything meaningful, same goes with the second example, Hints appreciated.

1. For the first one, it is easier to see the pattern if we write $$N = n^2$$. Then

$$\sum_{i=0}^{n^2} \frac{1}{7n^2 + i} = \sum_{i=0}^{N} \frac{1}{7N+i} = \sum_{i=0}^{N} \frac{1}{7+\frac{i}{N}} \cdot \frac{1}{N}.$$

As $$n \to \infty$$ (and hence $$N \to \infty$$), this converges to $$\int_{0}^{1} \frac{1}{7+x} \, \mathrm{d}x = \log(8/7)$$.

2. For the second one, we first substitute $$j = i - n$$ and then write $$N = 6n$$. Then

$$\sum_{i=n+1}^{7n} \frac{i}{n^2} = \sum_{j=1}^{6n} \frac{n + j}{n^2} = \sum_{j=1}^{N} \left( 1 + \frac{6j}{N} \right) \frac{6}{N}.$$

As $$n \to \infty$$, this converges to $$\int_{0}^{6} (1 + x) \, \mathrm{d}x = 24$$.

• In the first one I made a silly mistake and took $n$ as upper limit for $i$, while it should have been $n^2$ instead. Now it makes sense, thank You. Jul 6 at 7:04

The second sum, $$\sum_{i=n+1}^{7n} \frac{i}{n^2}$$ , does represent a Riemann sum for $$\int_1^7 x\,dx=24$$. Alternatively, we see that

\begin{align} \sum_{i=n+1}^{7n} \frac{i}{n^2}&=\sum_{i=0}^{7n} \frac{i}{n^2}-\sum_{i=0}^{n} \frac{i}{n^2}\\\\&=\frac12 (7n)(7n+1)-\frac12 n(n+1)\\\\ &=24 \end{align}

as expected!

• I think that the sum should run to $n^2$, not $n$. It seems numerically to converge to something larger than $0$.
– PC1
Jul 5 at 18:56
• I thought the same, but the sup limit for $k$ is $n^2$, not $n$ Jul 5 at 18:56
• @PC1 Yes. You are correct. Jul 5 at 19:09