# Infinite series

1. Nov 4, 2004

### kdinser

Today we started on infinite series, I'm getting the material just fine and able to do most of the problems, but one is giving me problems.

$$\lim_{n\to{a}} 2n/\sqrt{n^2+1}$$

I recognized that $$\infty/\infty$$ so I can use L'Hopital's rule. So taking the derivative of the numerator and denominator I get.

$$\frac{d}{dn} 2n = 2$$
and
$$\frac{d}{dn} \sqrt{n^2+1} = \frac{n}{\sqrt{n^2+1}}$$

Somehow the solutions manual is getting
$$2/\sqrt{1+(1/n^2)}$$

and a final answer of 2.

I can't see how they turned what I get for the derivative in the denominator into what they use. L'Hopital's rule twice would get rid of the n on top and put a 2 there, but that wouldn't change the square root.

thanks for any help

Last edited: Nov 4, 2004
2. Nov 4, 2004

### Galileo

Simply divide the numerator and the denominator by n. You get:
$$\frac{2n}{\sqrt{n^2+1}}=\frac{2}{\sqrt{1+1/n^2}}$$
I'd advise against using l'hospitals rule. Although it may be valid to use in some cases, you generally have to show it actually is.
In this case, n is restricted to integer values so the derivative doesn't exist.

3. Nov 4, 2004

### kdinser

Thanks for the help,

Could you go through a little of the algebra in the denominator, it's still not making sense to me how it's actually done. What property allows you to divide the radical by n and come up with that?

4. Nov 4, 2004

### kdinser

Doh, nevermind, it popped into my head as I was taking a shower. Thanks again.

5. Nov 4, 2004

### HallsofIvy

Staff Emeritus
By the way, in mathematics, there is a technical difference between "sequences" and "series". What you are dealing with here are "series", not "sequences".

6. Nov 4, 2004

### Galileo

Haha, not the other way around? :tongue2:
Don't confuse them indeed.