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.


2 Answers 2


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 $\int_{0}^{6} (1 + x) \, \mathrm{d}x = 24$.

  • $\begingroup$ 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. $\endgroup$
    – Kombajn
    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!

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

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