Finding limit using integration

[songoku]

Homework Statement

Please see below

Relevant Equations

Riemann Sum

I want to ask why the answer is not zero. If n approaches infinity, it means each term will approach zero so why the answer is not zero?

Thanks

 

[renormalize]

I want to ask why the answer is not zero. If n approaches infinity, it means each term will approach zero so why the answer is not zero?

True, each term does approach zero, but the number of terms in the sum approaches infinity. Can't the two effects compensate each other to yield a finite result?

 

[songoku]

True, each term does approach zero, but the number of terms in the sum approaches infinity. Can't the two effects compensate each other to yield a finite result?

I am thinking something like 0 + 0 + 0 + 0 + ..... = 0. Not sure how the compensation works

Thanks

 

[FactChecker]

I am thinking something like 0 + 0 + 0 + 0 + ..... = 0. Not sure how the compensation works

Consider this simpler example where each term goes to zero, but the sum of the n terms is always 1.
(n terms): 1/n+1/n+1/n+...+1/n = n/n = 1

 

[Mark44]

I want to ask why the answer is not zero.

Because that's not one of the listed potential answers.

I am thinking something like 0 + 0 + 0 + 0 + ..... = 0.

That's not reasonable -- you can't take the limit of each individual term.

Not sure how the compensation works

What is meant by compensation here is that although each term is approaching zero, there are more and more terms. The decreasing size of each term is compensated by the contrasting effect of adding more and more of them. Kind of like inflating a balloon that has a hole in it. If the amount of air escaping the balloon is more than the amount coming in, the balloon won't be inflated, but if more air is coming in than going out, the balloon inflates. The air coming in can compensate for the air going out.

 

[songoku]

That's not reasonable -- you can't take the limit of each individual term.

Is the reason that the number of terms is infinite?

Thanks

 

[Mark44]

Is the reason that the number of terms is infinite?

Yes.

 

[songoku]

Thank you very much for all the help and explanation renormalize, FactChecker, Mark44

 

[docnet]

@songoku if you already figured it out, can you post the solution for reference?

 

[songoku]

@songoku if you already figured it out, can you post the solution for reference?

$$\lim_{n\to \infty} \left(\frac{1}{2n+1}+\frac{1}{2n+2} + ... + \frac{1}{2n+n} \right)$$
$$=\lim_{n\to \infty} \sum_{r=1}^{n} \left(\frac{1}{2n+r}\right)$$
$$=\lim_{n\to \infty} \sum_{r=1}^{n} \left(\frac{1}{2+\frac{r}{n}} . \frac{1}{n} \right)$$
$$=\int_{0}^{1} \frac{1}{2+x} dx$$
$$=\ln(2+x)|^{1}_{0}$$
$$=\ln(3) - \ln(2)$$
$$=\ln \left(\frac{3}{2} \right)$$

 

[docnet]

$$=\lim_{n\to \infty} \sum_{r=1}^{n} \left(\frac{1}{2+\frac{r}{n}} . \frac{1}{n} \right)$$
$$=\int_{0}^{1} \frac{1}{2+x} dx$$

This step is not immediately clear to me. Could someone explain why the limit is the integral?

 

[songoku]

This step is not immediately clear to me. Could someone explain why the limit is the integral?

For me, I imagine the rectangles of Riemann sum

 

[docnet]

It must be as $x=\frac{r}{n}$ and $\Delta x = \frac{1}{n}$. The boundaries are $0$ and $1$ because
$$0< r \leq n\Leftrightarrow 0<\frac{r}{n}\leq 1\Leftrightarrow 0<x\leq 1,$$
which goes along with $$\sum_{r=1}^n\frac{1}{n}=1$$
for the traditional right Riemann sum $$\sum_{r=1}^n \frac{1}{2+x_r}\Delta x\,$$
and $(x_0,...,x_n)$ an equally spaced partition of $[0,1].$

 

[pasmith]

If $f : [0, \infty) \to \mathbb{R}$ is strictly decreasing, we have $$\int_1^{n+1} f(x)\,dx \leq \sum_{r=1}^n f(r) \leq \int_0^n f(x)\,dx$$ which for $f(x) = (2n + x)^{-1}$ is $$\ln \left( \frac{3n + 1}{2n + 1} \right) \leq \sum_{r=1}^n \frac{1}{2n + r} \leq \ln \tfrac 32.$$ Taking the limit $n \to \infty$ establishes the result.

 

[Gavran]

$$=\lim_{n\to \infty} \sum_{r=1}^{n} \left(\frac{1}{2+\frac{r}{n}} . \frac{1}{n} \right)$$
$$=\int_{0}^{1} \frac{1}{2+x} dx$$

$$ \int_{0}^{1}\frac{1}{2+x}dx \approx \sum_{r=1}^{n}\left(\frac{1}{2+\frac{r}{n}}\cdot\frac{1}{n}\right) = \sum_{r=1}^{n}\left(\frac{1}{2+(0+r\frac{1-0}{n})}\cdot\frac{1-0}{n}\right) $$ is a right rectangular approximation or a right Riemann sum.