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

In summary: So the sum of the terms with ##\sqrt{n+1} + \sqrt{n-1}## is smaller than the sum of the terms with ##\sqrt{2n}##, and we can see the same thing from the partial sums, and so on.In summary, we can determine which of the two infinite series is larger by comparing the terms with the same square root and using the fact that the function ##x^2## is strictly increasing for ##x \gt 0##.
  • #1
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Homework Statement
No homework.. I dont know why was moved...
Relevant Equations
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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]
 
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  • #2
I realized this mistake, during using converging test... How should I know which is bigger?
 
  • #3
Making the ratio of the terms
[tex]\frac{1}{2}(\sqrt{1+ 1/n}+\sqrt{1- 1/n}\ )=\frac{1}{2}(2- k/n^2+...) < 1 [/tex]
k is a positive constant that you can get by Taylor expansion.
 
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  • #4
anuttarasammyak said:
Making ratio of the terms
[tex]\frac{1}{2}(\sqrt{1+ 1/n}+\sqrt{1- 1/n}\ )=\frac{1}{2}(2- k/n^2+...) < 1 [/tex]
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.
 
  • #5
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...
 
  • #6
You see

[tex](\sqrt{1+1/n}+\sqrt{1-1/n})^2=2+2\sqrt{1+1/n}\sqrt{1-1/n}<4[/tex]
[tex]\sqrt{1+1/n}+\sqrt{1-1/n}<2[/tex]
 
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  • #7
anuttarasammyak said:
I correct the mistake of signature in #3. You see

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

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..
 
  • #8
I corrected my mistake and edit #6. I hope you get it.
 
  • #9
anuttarasammyak said:
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.
 
  • #10
For n-1
[tex]\sqrt{1+1/(n-1)}+\sqrt{1-1/(n-1)}<2[/tex]
For n+1
[tex]\sqrt{1+1/(n+1)}+\sqrt{1-1/(n+1)}<2[/tex]
Same things.
 
  • #11
anuttarasammyak said:
For n-1
[tex]\sqrt{1+1/(n-1)}+\sqrt{1-1/(n-1)}<2[/tex]
For n+1
[tex]\sqrt{1+1/(n+1)}+\sqrt{1-1/(n+1)}<2[/tex]
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 $^-1$
 
  • #12
physics1000 said:
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}##
physics1000 said:
If I will understand this, it will be easier for me to understand, so then I just do $^-1$
 
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  • #13
Mark44 said:
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 :)
 
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  • #14
physics1000 said:
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.
 
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  • #15
Going back to post #1:
physics1000 said:
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}##.
 
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  • #16
FactChecker said:
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 :)
 
  • #17
Mark44 said:
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 :)
 

1. How do I compare two infinite series?

To compare two infinite series, you need to look at their terms and see if they follow a pattern. If the terms of one series are always larger than the corresponding terms of the other series, then the first series is bigger. If the terms of one series are always smaller than the corresponding terms of the other series, then the second series is bigger. If the terms of both series follow a similar pattern, then you will need to use a convergence test to determine which series is bigger.

2. What is a convergence test?

A convergence test is a mathematical tool used to determine if an infinite series converges or diverges. There are various convergence tests, such as the comparison test, ratio test, and integral test, which can help you determine which series is bigger. These tests involve comparing the terms of the series to a known convergent or divergent series.

3. Can I compare two infinite series with different starting points?

Yes, you can compare two infinite series with different starting points. However, you will need to take into account the starting point when using a convergence test. For example, if one series starts at n=0 and the other starts at n=1, you will need to adjust the test accordingly. It is important to note that the starting point does not affect the overall comparison of which series is bigger.

4. How do I know if an infinite series is bigger than another?

If the terms of one series are always larger than the corresponding terms of the other series, then the first series is bigger. If the terms of one series are always smaller than the corresponding terms of the other series, then the second series is bigger. If the terms of both series follow a similar pattern, then you will need to use a convergence test to determine which series is bigger.

5. Is it possible for two infinite series to be equal in size?

Yes, it is possible for two infinite series to be equal in size. This occurs when the terms of both series follow the exact same pattern, and both series converge or diverge at the same rate. However, it is more common for one series to be bigger than the other, as the terms of infinite series can vary greatly in size.

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