1
$\begingroup$

Find $\lim \limits_{n \to \infty} (a_n)$, where $a_n=\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+...+\frac{n}{n^2}$.

So I can solve it like that $a_n=\frac{1+2+3+...n}{n^2}=\frac{\frac{1}{2}n(n+1)}{n^2}=\frac{1}{2}(1+\frac{1}{n})$. Clearly $\lim \limits_{n \to \infty} (a_n)=\frac{1}{2}$ so the sequence converges to $\frac{1}{2}$.

But can I approximate the series as an integral? $$a_n=\sum_{i=1}^n \frac{i}{n^2} \approx\int_{1}^n \frac{x}{n^2}dx=\frac{1}{n^2}\int_{1}^n x \,dx=\frac{1}{2}-\frac{1}{2n^2}$$

Now, when $n$ tends to infinity, $a_n$ tends to $\frac{1}{2}$ so the sequence converges to $\frac{1}{2}$. This produced the same result as using the first method. The only thing I am unsure of is that the final sums are different despite the fact that they both converge to the same number. This is because in the first method we sum only integers but in the second we sum all real $x$'s in the given interval, right?

Is this approach also valid?

$\endgroup$
0
$\begingroup$

That approach is valid IF instead of approximating, you write either

$$ a_n=\sum_{i=1}^n \frac{i}{n^2} \le \int_{A_1}^{A_2} \frac{x}{n^2}dx=\ldots $$or

$$ a_n=\sum_{i=1}^n \frac{i}{n^2} \ge \int_{B_1}^{B_2} \frac{x}{n^2}dx=\ldots $$ where you have to choose the limits carefully. In the first case, you'd choose $A_1 = 1, A_2 = n+1$, so that on each interval, you've have $$ \frac{i}{n^2} \le \int_i^{i+1} \frac{x}{n^2} dx $$ i.e. $$ i \le \int_i^{i+1} x~ dx $$ which is valid because on the interval from $i \le x \le i+1$, the number $i$ really is no more than $x$.

In the second case (which you'd use if you were trying to prove divergence), you'd have to pick $B_1 = 0, B_2 = n$, and then observe that for $$ i-1 \le x \le i,$$ you have $x \le i$, and hence $$ i \ge \int_{i-1}^i x ~dx. $$

In both cases, it really helps that the function you're integrating ($x$ in this case) is monotone on each interval, so that its integral over the interval can be estimated (either a lower or upper bound estimate) by its value at one of the interval's endpoints.

You might try to perform a similar estimate to check the convergence/divergence of $a_n = \sin n$; you'll find that without the monotone-ness, you really can't get anywhere.

General note: saying that something is "approximately" something else is generally a risky business. For instance, in the sequences $a_n = 1/n$ and $b_n = 1/n^2$, the later terms are approximately zero...but one diverges and the other converges. Far better to have definitive upper or lower bounds whenever possible.

$\endgroup$
0
$\begingroup$

This can be written as $$\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^n {r\over n}$$ This is of the form $$\lim_{n \to \infty}\frac{1}{n} \sum_{r=1}^n f\left ({r\over n} \right) $$ So it can be written as $$\int_0^1 f(x)dx$$ $$=\int_0^1x dx$$ $$=\frac{1}{2}$$

$\endgroup$
  • $\begingroup$ That's true, but it's not an answer to the question that OP asked, which was whether comparing to a particular integral was a valid way to test convergence of a series. $\endgroup$ – John Hughes Apr 9 '16 at 12:34
  • $\begingroup$ Isn't he asking a way to do this using integrals? $\endgroup$ – Nikunj Apr 9 '16 at 12:44
  • $\begingroup$ @Nikunj - Thanks for the answer, but why can you write this as an integral from 0 to 1? I think I'm missing something. $\endgroup$ – Richard Smith Apr 9 '16 at 18:18
  • $\begingroup$ I think he's asking "is this approach also valid?", where "this approach" is to approximate a partial sum of the series as an integral over the interval $1 \le x \le n$ (or perhaps $n+1$). You've shown that each partial sum is also a "riemann sum" for an integral of a different function on the interval $[0, 1]$. This seem to me to be rather different ways to use integrals to estimate a partial sum, although they're clearly related. No matter -- if OP learns something for your answer, then it's a good one. :) $\endgroup$ – John Hughes Apr 9 '16 at 19:34
  • $\begingroup$ @RichardSmith while $r$ takes discrete values {$0,1,2...$} $\frac{r}{n}$ can take any value between $0$ to $1$ $\endgroup$ – Nikunj Apr 9 '16 at 20:07
0
$\begingroup$

The integral approach is valid because

(1) The function $\;f(x)=x\;$ is integrable in any finite interval (for example, because it is everywhere continuous), and

(2) The sum is a very specific Riemann sum of the above function in the interval $\;[0,1]\;$ with respect to a very specific partition of this interval and choosing very specific points within each subinterval in that partition..

By (1), any Riemann sum with respect to any partition will converge to the integral $\;\int_0^1 x\,dx\;$ .

This is just like choosing a very specific subsequence of a sequence that we know beforehand that it is convergent: the subsequence is going to converge and to the same limit that the whole function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.