How to know which is bigger? (Comparing two infinite series)

[physics1000]

Homework Statement

No homework.. I dont know why was moved...

Relevant Equations

.

Summary:: How to know which one is bigger when n goes to infinity?
$$ \sum_{n=1}^\infty \frac {1} {\sqrt {n}(\sqrt {n+1}+\sqrt {n-1})} $$
And:
$$ \sum_{n=1}^\infty \frac {1} {\sqrt {n}(\sqrt {n}+\sqrt {n})} $$

I thought at first that the second one is bigger, although, I came to realize, to my mistake, that the first one is actually bigger.
How do I know which is the bigger one at numbers like those?

EDIT:
You can think of it like that:
$$\sqrt {n-1}+\sqrt{n+1} < \sqrt{2n}$$
Why is it like that? That is my problem to understand
[Mentor Note -- Thread has been moved from the technical forums to the schoolwork forums]

 

[physics1000]

I realized this mistake, during using converging test... How should I know which is bigger?

 

[anuttarasammyak]

Making the ratio of the terms
$$\frac{1}{2}(\sqrt{1+ 1/n}+\sqrt{1- 1/n}\ )=\frac{1}{2}(2- k/n^2+...) < 1$$
k is a positive constant that you can get by Taylor expansion.

 

[physics1000]

Making ratio of the terms
$$\frac{1}{2}(\sqrt{1+ 1/n}+\sqrt{1- 1/n}\ )=\frac{1}{2}(2- k/n^2+...) < 1$$
k is a positive constant that you can get by Taylor expansion.

Sadly, we have not learned Taylor yet, so we are not allowed to use it.
only basic stuff you learn in Semester one at calculus.

 

[physics1000]

But anyway, I actually just found out ( by using calculator ) it is always like that, regardless of n.
Why is it like that? Is there a way to find out stuff like that by using basic means?
I saw even if its n+2 and n-2, its also, and such...

 

[anuttarasammyak]

You see

$$(\sqrt{1+1/n}+\sqrt{1-1/n})^2=2+2\sqrt{1+1/n}\sqrt{1-1/n}<4$$
$$\sqrt{1+1/n}+\sqrt{1-1/n}<2$$

 

[physics1000]

I correct the mistake of signature in #3. You see

$$(\sqrt{1+1/n}+\sqrt{1-1/n})^2=4+2\sqrt{1+1/n}\sqrt{1-1/n}>2^2$$
$$\sqrt{1+1/n}+\sqrt{1-1/n}>2$$

Ahh, I can't understand this math.. its too complicated for me...
But regardless of it, how does it relate to my question? I really can't understand... you did something else, I did n+1 n-1, you did n\1..

 

[anuttarasammyak]

I corrected my mistake and edit #6. I hope you get it.

 

[physics1000]

I corrected my mistake and edit #6. I hope you get it.

I will get it once I understand something.
How does it relate to my question? the fact you did 1/n and not n-1 or n+1.
That is what troubling me, I can't understand why you chose that especially.

 

[anuttarasammyak]

For n-1
$$\sqrt{1+1/(n-1)}+\sqrt{1-1/(n-1)}<2$$
For n+1
$$\sqrt{1+1/(n+1)}+\sqrt{1-1/(n+1)}<2$$
Same things.

 

[physics1000]

For n-1
$$\sqrt{1+1/(n-1)}+\sqrt{1-1/(n-1)}<2$$
For n+1
$$\sqrt{1+1/(n+1)}+\sqrt{1-1/(n+1)}<2$$
Same things.

Maybe I will explain myself better, I can actually change the subject so it will be easier:
Why is:
$$\sqrt {n-1}+\sqrt{n+1} < \sqrt{2n}$$
If I will understand this, it will be easier for me to understand, so then I just do $^-1$

 

[Mark44]

Maybe I will explain myself better, I can actually change the subject so it will be easier:
Why is:
$$\sqrt {n-1}+\sqrt{n+1} < \sqrt{2n}$$

This is not true.

Suppose for the moment that this inequality is true.
$\sqrt {n-1}+\sqrt{n+1} < \sqrt{2n}$
$\Rightarrow n - 1 + 2\sqrt{n - 1}\sqrt{n + 1} + n + 1 < 2n$ (squaring both sides of the above inequality)
$\Rightarrow 2n + 2\sqrt{n^2 - 1} < 2n$
This is a contradiction, since $2\sqrt{n^2 - 1} > 0$ for n > 1. 2n + a positive value can't be less than 2n.
This means that my assumption in the first line above must be false. Hence $\sqrt {n-1}+\sqrt{n+1} \ge \sqrt{2n}$

If I will understand this, it will be easier for me to understand, so then I just do $^-1$

 

[physics1000]

This is not true.

Suppose for the moment that this inequality is true.
$\sqrt {n-1}+\sqrt{n+1} < \sqrt{2n}$
$\Rightarrow n - 1 + 2\sqrt{n - 1}\sqrt{n + 1} + n + 1 < 2n$ (squaring both sides of the above inequality)
$\Rightarrow 2n + 2\sqrt{n^2 - 1} < 2n$
This is a contradiction, since $2\sqrt{n^2 - 1} > 0$ for n > 1. 2n + a positive value can't be less than 2n.
This means that my assumption in the first line above must be false. Hence $\sqrt {n-1}+\sqrt{n+1} \ge \sqrt{2n}$

Oh dam, my bad, I meant 2sqrt(n), not sqrt(2n), sorry..
And I just managed to understand it, thanks :) ( Asked a friend ). But your prove also helped me certificate it. Thanks!
post can be closed if needed :)

 

[FactChecker]

Oh dam, my bad, I meant 2sqrt(n), not sqrt(2n), sorry..
And I just managed to understand it, thanks :) ( Asked a friend ). But your prove also helped me certificate it. Thanks!
post can be closed if needed :)

Just to complete the explanation using only algebra:
Because the function $x^2$ is strictly increasing for $x \gt 0$, $\sqrt{n+1} + \sqrt{n-1} \lt 2\sqrt{n}$ if and only if $(\sqrt{n+1} + \sqrt{n-1} )^2\lt (2\sqrt{n})^2$
The right hand side is $4n$ and the left hand side is
$(n+1) + 2\sqrt{n+1}\sqrt{n-1} + (n-1)$
$= 2n+2\sqrt{n^2-1}$
$\lt 2n+2\sqrt{n^2}$ (because $\sqrt{x}$ is strictly increasing for $x\gt 0$.)
$= 4n$

Since the denominators of the terms with $\sqrt{n+1} + \sqrt{n-1}$ are smaller, those terms are all larger.

 

[Mark44]

Going back to post #1:

Summary:: How to know which one is bigger when n goes to infinity?
$$ \sum_{n=1}^\infty \frac {1} {\sqrt {n}(\sqrt {n+1}+\sqrt {n-1})} $$
And:
$$ \sum_{n=1}^\infty \frac {1} {\sqrt {n}(\sqrt {n}+\sqrt {n})} $$

I thought at first that the second one is bigger, although, I came to realize, to my mistake, that the first one is actually bigger.
How do I know which is the bigger one at numbers like those?

The second series can be rewritten as $\sum_{n=1}^\infty \frac {1} {2n}$, which behaves exactly the same as the harmonic series, $\sum \frac 1 n$. IOW, both series diverge.

Since you ask about which series is bigger, you are apparently using the comparison test. If you believe that the first series also diverges, you need to show that each term of the first series is equal to or larger than the corresponding term of the series $\sum \frac 1 {2n}$. Because it has been shown that each term in the first series is smaller than 2n, this comparison doesn't tell you anything at all.

On the other hand, if you believe that the first series diverges, you need to do a comparison with a different series -- one that is known to converge, and you need to show that each term of your series is equal to or smaller than the corresponding term of the known series.

The limit comparison test would be a better choice to determine the behavior of the series you're investigating, again comparing it with the series $\sum \frac1 {2n}$.

 

[physics1000]

Just to complete the explanation using only algebra:
Because the function $x^2$ is strictly increasing for $x \gt 0$, $\sqrt{n+1} + \sqrt{n-1} \lt 2\sqrt{n}$ if and only if $(\sqrt{n+1} + \sqrt{n-1} )^2\lt (2\sqrt{n})^2$
The right hand side is $4n$ and the left hand side is
$(n+1) + 2\sqrt{n+1}\sqrt{n-1} + (n-1)$
$= 2n+2\sqrt{n^2-1}$
$\lt 2n+2\sqrt{n^2}$ (because $\sqrt{x}$ is strictly increasing for $x\gt 0$.)
$= 4n$

Since the denominators of the terms with $\sqrt{n+1} + \sqrt{n-1}$ are smaller, those terms are all larger.

Thanks for the explanation :)

 

[physics1000]

Going back to post #1:

The second series can be rewritten as $\sum_{n=1}^\infty \frac {1} {2n}$, which behaves exactly the same as the harmonic series, $\sum \frac 1 n$. IOW, both series diverge.

Since you ask about which series is bigger, you are apparently using the comparison test. If you believe that the first series also diverges, you need to show that each term of the first series is equal to or larger than the corresponding term of the series $\sum \frac 1 {2n}$. Because it has been shown that each term in the first series is smaller than 2n, this comparison doesn't tell you anything at all.

On the other hand, if you believe that the first series diverges, you need to do a comparison with a different series -- one that is known to converge, and you need to show that each term of your series is equal to or smaller than the corresponding term of the known series.

The limit comparison test would be a better choice to determine the behavior of the series you're investigating, again comparing it with the series $\sum \frac1 {2n}$.

Oh I managed to answer it. the original question was something else. I actually used the limit as you said. showed that my original is bigger then 1\n and thus it diverges.
What I wrote here, was something close to what I needed, when I realized it was sqrt(n+1)+sqrt(n-1)<2sqrt(n), it was easier, I was just a little bit idiotic in thinking that 2sqrt(n) is smaller then that hehe.
Thanks :)