# Monotonous functions and asymptotes

1. Mar 10, 2007

### Dragonfall

1. The problem statement, all variables and given/known data

Show that if $$f(x)$$ tends to zero monotonically as x increases without limit, and is continuous for x>0, and of the series $$\sum_{k=1}^{\infty}f(k)$$ diverges, then $$\sum_{k=1}^nf(k) \sim \int_1^nf(x)dx$$.

If g(x) is a second function satisfying the same hypotheses as f(x), and if g(x)=o(f(x)), show that

$$\sum_{k=1}^ng(k)=o(\sum_{k=1}^nf(k)_$$.

3. The attempt at a solution

I think by monotonicity we can conclude that

$$\frac{\int_1^nf(x+1)dx}{\sum_{k=1}^nf(k)}\le1\le\frac{f(1)+\int_1^nf(x)dx}{\sum_{k=1}^nf(k)}$$

And I want to show that LHS and RHS converge to each other, hence they both converge to 1. But I can't get it to work.

The second part I don't know what to do.

Last edited: Mar 10, 2007
2. Mar 10, 2007

### gammamcc

I think you have to restate the problem. How does n enter the picture?

3. Mar 10, 2007

### lalbatros

Dragonfall,

I have two questions to understand your question:

Do you really mean that $$\sum_{k=1}^{\infty}f(k)$$ diverges ?

What do you mean by $$\sum_{k=1}^\infty f(k) \sim \int_1^nf(x)dx$$, what is the meaning of the symbol $$\sim$$ ?

Otherwise I have the feeling that a correspondance between a series and an integral can always be established by considering some ad-hoc step function. The properties could then be derived by considering some relations between the step function and the continuous function. In your case, with the given hypothesis, you can easily find two step functions majoring and minoring the integral.

michel

Last edited: Mar 10, 2007
4. Mar 10, 2007

### Gib Z

$$\sim$$ Means the quotient of the two, in the limit of infinity, is equal to 1. Note: Just because the quotient of the two is 1, it does not mean they are equal, even in the limit.

Eg $$x \sim x+1$$, but the difference is always constant and they are never equal.

In long terms, hes saying

$$\lim_{n\to\infty} \frac{\sum_{k=1}^\infty f(k)}{\int_1^nf(x)dx}$$

5. Mar 10, 2007

### ZioX

Are you sure the limiting variable is n? n Doesn't appear in that quotient, hence it is a constant.

The problem needs to be restated.

6. Mar 10, 2007

### Gib Z

Yes it does appear in the quotient, look at the bounds on the integral..

7. Mar 10, 2007

### ZioX

I meant the numerator. My bad.

8. Mar 10, 2007

### Gib Z

No its just that before the original sum was $$\sum_{k=1}^n f(k)$$, and the limit makes that to infinity.

9. Mar 10, 2007

### Gib Z

If it make anyone happy:

$$\frac{\sum_{k=1}^\infty f(k)}{\int_1^{\infty} f(x) dx}$$

10. Mar 10, 2007

### ZioX

Hum!

$$\sum_{k=1}^{\infty}f(k)$$ doesn't even exist according to our hypothesis. Someone is being sloppy with notation.

11. Mar 10, 2007

### Gib Z

It wasn't me! I was just copying down whatever he said and explained what the ~ meant!

12. Mar 10, 2007

### HallsofIvy

Staff Emeritus
Perhaps it would be better to say
$$\lim_{n\rightarrow \infty}\frac{\Sigma_{k=1}^n f(k)}{\int_1^n f(x)dx}= 1$$
I believe that's the same thing.

Dragonfall, consider how the Riemann sum
$$\sum_{i=1}^n f(a+ i\Delta n)\Delta n[/itex] a an integer compares with f((a). 13. Mar 10, 2007 ### Dragonfall I'm not sure it's necessary to look at that sum, since f(x) is monotone decreasing we already have that [tex]\sum_{k=1}^nf(k)\le\int_1^nf(x)dx+f(1)$$,

and a similar bound from below. My question is whether

$$\frac{\int_1^nf(x+1)dx}{\sum_{k=1}^nf(k)}\rightarrow\frac{f(1)+\int_1^nf(x)dx}{\sum_{k=1}^nf(k)}$$

as n goes to infty, and second if

$$f(x)\le a\le g(x)$$

for all x and that as x increases $$f\rightarrow g$$ and $$g\rightarrow f$$, then $$f\rightarrow a$$?

Last edited: Mar 10, 2007