# Preparing for test

I have two questions i hope you can help me with:

The first one is about series of positive terms like, for example $$\sum _{i=0}^{inf}(n^{1/n}-1)$$.
Can i say that as the square of this series of positive terms diverges by the "nth term test", the original series also diverges.
If i can't use this, could you please point me on the right direction in order to do determine the convergence of this series?.

My second question is about a limit by definition(lambda-epsilon proof). I have to find out if the following function is continuous and/or differentiable:
$$f(x,y) =\left\{\begin{array}{cc}\frac {x*y} {\sqrt x -\sqrt y},&\mbox{ if }x\neq y\\0, & \mbox{x=y}\end{array}\right$$

Everything i tried indicates that this function is continuous, but i can't prove that the limit when (x,y)->(0,0) of f is in fact 0.
I you want me to explain a little better what i have done just ask. I would really appreciate any help you could give me.
Many Thanks, Paul.

Homework Helper
A general hint for these kind of limit problems. When you want to prove such a limit exists and is zero, you usually need to find an upper-bound for that function which also tends to zero. This is basically the squeeze theorem.
There are probably other ways to do it, but playing around I found:

$$\left| \frac{xy}{\sqrt{x}-\sqrt{y}}\right| \leq \frac{|xy|}{\sqrt{x}+\sqrt{y}}= |x|\cdot \left| \frac{y}{\sqrt{x}+\sqrt{y}}\right| \leq |x|\sqrt{y}$$
The first inequality uses the triangle inequality on the denominator. The second is obvious.
The right side clearly goes to zero.

thanks

Ok, I understand what you did, and i was thinking to do something very similar, but the inequalities didnt came up to my mind. I now have to find out if it is differentiable or not. Would i be on the right track if i use the theorem that says that if the partial derivatives are continuous in an "interval", then the function is also differentiable on that "interval"? I think i should use this instead of the differentiability limit, because i don't know which are the points to test for differentiability. Am i right?

Well i found out the partial derivatives and i realized that they are continuous everywhere except on (0,0) (I forgot to tell on the initial problem statement that the function is defined only for x and y postive or equal to 0), so i used the differentiability limit and i found that it was differentiable on (0,0).
Is my answer and PROCEDURE correct?

Do you have any hints for the first problem?

Thankfully, Paul.

Homework Helper
Galileo said:
$$\left| \frac{xy}{\sqrt{x}-\sqrt{y}}\right| \leq \frac{|xy|}{\sqrt{x}+\sqrt{y}}$$
This is a mistake. The right side is smaller than the left side, since $\sqrt{x}+\sqrt{y} \geq |\sqrt{x}-\sqrt{y}|$

checking

OOOUch, i didnt check your work correctly the first time, but now i see it was wrong. So then how could i prove this limit?
Assuming i could prove the continuity limit, i am pretty sure i could prove the differntiability also.
Grateful, Paul.

iNCREDiBLE
pbialos said:
OOOUch, i didnt check your work correctly the first time, but now i see it was wrong. So then how could i prove this limit?
Assuming i could prove the continuity limit, i am pretty sure i could prove the differntiability also.
Grateful, Paul.

Polar Coordinates.

Homework Helper
Actually, it's pretty easy to see the limit doesn't exist (so no need to check for differentiability).

Note that the line y=x (where x>0) doesn't lie in the domain of f. The denominator can gets arbitrality small in any neighbourhood around (0,0). Therefore, if you choose an $\epsilon$-neighbourhood around (0,0), no matter how small, you can approach the line y=x from two sides inside that neighbourhood. Take a circle of radius $r<\epsilon$ and approach y=x from two sides. The product xy is nonzero near that line, but $\sqrt{x}-\sqrt{y}$ approaches $0^+$ and $0^-$ respectively.

About the sum, you use 'i' as your index, but have 'n' in the expression, did you mean:

$$\sum_{n=0}^{\infty}(n^{1/n}-1)$$

I`m not familiar with the 'n'th-term test'.

ok

ok, so problem solved for the second one.
About the first one, it is as you said:
$$\sum_{n=0}^{\infty}(n^{1/n}-1)$$

And what i called the "n-th term root" is just the propertie that says that if a series $$\sum_{n=0}^{\infty}(A_n)$$ converges, then $$\lim_{n->\infty}A_n=0$$

I am sorry, but i didnt know how this property was called in english.

I found another way to solve this exercise but again i am not sure if what i did is correct:
I divided $$n^{1/n}-1$$ by $$n^{1/n}$$ and took the limit of this when n tends to infinite. It gave me 0, so i concluded that the series:
$$\sum_{n=0}^{\infty}(n^{1/n}-1)$$ and $$\sum_{n=0}^{\infty}(n^{1/n})$$ behave the same way(i think it is because the limit comparison test right??).
As the second series diverges, then the original series does also diverges.

I would really appreciate if you could tell me if any of the two ways i tried are correct.
I have my doubts about the second one specially, because i don't remember if i could use the limit comparison test when the limit A_n/B_n is equal to 0 with n tending to infinite.

Thank you, Paul.

Last edited by a moderator: