Series Test Question

[slydez]

Hi

I'm new here and was looking to get some help on my calculus homework. I need a confirmation on the Test for Divergent for series. So if the limit of a series is infinity does it diverge by Test for Divergent?


I also need help on this problem.

n=1 to infinity of the sum of (sqrt(n+2)-2sqrt(n+1)+sqrt(n))

How do I even start? I tried using limit and when I manipulated it I ended with 0/0. Thus, I asked the previous question.


Thank you


Edit:

Sorry, I replied like 5 times 5 hours ago and it seems that it still hasn't been reviewed or something. Is it always like this in this website?
Anyways I'll just reply through edits here.

What I did was multiplying the series equation to (-sqrt(n+2)+2sqrt(n+1)-sqrt(n)) / (-sqrt(n+2)+2sqrt(n+1)-sqrt(n)). Then I factored out sqrt(n^2) on both the numerator and denominator. Then I found that the limit of it equals to 0/0.


Step 1: (sqrt(n+2)-2sqrt(n+1)+sqrt(n)) x [(-sqrt(n+2)+2sqrt(n+1)-sqrt(n)) / (-sqrt(n+2)+2sqrt(n+1)-sqrt(n))]

Step 2: [-6n+4sqrt(n^2+3n+2)-2sqrt(n^2+2n)+4sqrt(n^2+n)-3] / [-sqrt(n+2)+2sqrt(n+1)-sqrt(n)]

Step 3: [sqrt(n^2) (-6+4sqrt(1+3/n+2/n^2)-2sqrt(1+2/n)+4sqrt(1+1/n)-3/n)] / [sqrt(n^2) (-sqrt(1/n+2/n^2)+2sqrt(1/n+1/n^2)-sqrt(1/n))]

Step 4: lim of the above equation as n approaches infinity = 0/0 Which means DNE? Therefore it is diverges by Test for Divergence? <--This is the part where I'm wondering if it is correct to apply it here.


Is there a way to show my work better? I feel that typing it out looks messy. Is there a coding outline here to use for math symbols?


Edit 2:

Write out some of the first expressions in staggered fashion and see if there if there is a telescoping sum.

Oh didn't think of that.
Okay, I got 1-sqrt(2)-sqrt(n+1)+sqrt(n+2) in which the limit as n approaches 0 is infinity. Does this mean it is divergent by Telescoping Series?

If so can Test for Divergence still work which is shown on my original work.

 

[Deleted member 4993]

Hi

I'm new here and was looking to get some help on my calculus homework. I need a confirmation on the Test for Divergent for series. So if the limit of a series is infinity does it diverge by Test for Divergent?


I also need help on this problem.

n=1 to infinity of the sum of (sqrt(n+2)-2sqrt(n+1)+sqrt(n))

How do I even start? I tried using limit and when I manipulated it I ended with 0/0. Thus, I asked the previous question.


Thank you

How did you get the division in there to get 0/0?

 

[slydez]

How did you get the division in there to get 0/0?

I multiplied it by (-sqrt(n+2)+2sqrt(n+1)-sqrt(n))/(-sqrt(n+2)+2sqrt(n+1)-sqrt(n)) then I factored out sqrt(n^2).




Is there a specific type of coding to make the proper signs to show the math better? Like sqrt turns into the actual sqrt.

 

[slydez]

How did you get the division in there to get 0/0?

I multiplied (-sqrt(n+2)+2sqrt(n+1)-sqrt(n))/(-sqrt(n+2)+2sqrt(n+1)-sqrt(n)) to the equation and then factored out sqrt(n^2). Which I ended up with 0/0.



Is there a math symbol coding guide? To use to make it look better?

 

[slydez]

Test...not sure if posting in this website takes a while...

Deleting this post if it does.

 

[Steven G]

Hi

I'm new here and was looking to get some help on my calculus homework. I need a confirmation on the Test for Divergent for series. So if the limit of a series is infinity does it diverge by Test for Divergent?


I also need help on this problem.

n=1 to infinity of the sum of (sqrt(n+2)-2sqrt(n+1)+sqrt(n))

How do I even start? I tried using limit and when I manipulated it I ended with 0/0. Thus, I asked the previous question.


Thank you

Where do you see a denominator to get 0/0?? Please show us your work.

 

[slydez]

I'm not sure but I think replying takes a while?

 

[lookagain]

n=1 to infinity of the sum of (sqrt(n+2)-2sqrt(n+1)+sqrt(n))

How do I even start?

Write out some of the first expressions in staggered fashion and see if there if there is a telescoping sum.


\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{3} \ - \ 2\sqrt{2} \ + \ 1\sqrt{1}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{4} \ - \ 2\sqrt{3} \ + \ 1\sqrt{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{5} \ - \ 2\sqrt{4} \ + \ 1\sqrt{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{6} \ - \ 2\sqrt{5} \ + \ 1\sqrt{4}\)
\(\displaystyle \ \ \ \ \ \ \ \ +1\sqrt{7} \ - \ 2\sqrt{6} \ + \ 1\sqrt{5}\)

and so on.




Add the columns. Which ones do not "cancel out"/sum to zero?

 

[slydez]

Write out some of the first expressions in staggered fashion and see if there if there is a telescoping sum.


\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{3} \ - \ 2\sqrt{2} \ + \ 1\sqrt{1}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{4} \ - \ 2\sqrt{3} \ + \ 1\sqrt{2}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{5} \ - \ 2\sqrt{4} \ + \ 1\sqrt{3}\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +1\sqrt{6} \ - \ 2\sqrt{5} \ + \ 1\sqrt{4}\)
\(\displaystyle \ \ \ \ \ \ \ \ +1\sqrt{7} \ - \ 2\sqrt{6} \ + \ 1\sqrt{5}\)

and so on.




Add the columns. Which ones do not "cancel out"/sum to zero?

Ohh didn't think of that.
Okay so I got 1-sqrt(2)-sqrt(n+1)+sqrt(n+2) which as n approaches 0 is infinity...Does that mean it is divergent by Telescoping Series?

If it does can the Test for Divergence still apply from my original work?

 

[Ishuda]

Ohh didn't think of that.
Okay so I got 1-sqrt(2)-sqrt(n+1)+sqrt(n+2) which as n approaches 0 is infinity...Does that mean it is divergent by Telescoping Series?

If it does can the Test for Divergence still apply from my original work?

Look at
\(\displaystyle \sqrt{n+2}\, -\, \sqrt{n+1}\)
part of the expression. As n goes to infinity that goes to \(\displaystyle \infty-\infty\) which may or may not be finite. Suppose you were to do the same sort of thing you did before and multiply both numerator and denominator by (n+2)^1/2+(n+1)^1/2 to get
\(\displaystyle \frac{\sqrt{n+2}\, -\, \sqrt{n+1}}{1}\, *\, \frac{\sqrt{n+2}\, +\, \sqrt{n+1}}{\sqrt{n+2}\, +\, \sqrt{n+1}}\)

 

[slydez]

Look at
\(\displaystyle \sqrt{n+2}\, -\, \sqrt{n+1}\)
part of the expression. As n goes to infinity that goes to \(\displaystyle \infty-\infty\) which may or may not be finite. Suppose you were to do the same sort of thing you did before and multiply both numerator and denominator by (n+2)^1/2+(n+1)^1/2 to get
\(\displaystyle \frac{\sqrt{n+2}\, -\, \sqrt{n+1}}{1}\, *\, \frac{\sqrt{n+2}\, +\, \sqrt{n+1}}{\sqrt{n+2}\, +\, \sqrt{n+1}}\)

I got

\(\displaystyle \frac{-1-1/n\, +\, 2\sqrt{1+3/n+2/n^2}\, -\, 1-2/n}{\sqrt{1/n+1/n^2}\, -\, \sqrt{1/n+2/n^2}}\)

In which the limit is 0/0...it is still undefined...
Is it suppose to converge?

 

[Ishuda]

I got

\(\displaystyle \frac{-1-1/n\, +\, 2\sqrt{1+3/n+2/n^2}\, -\, 1-2/n}{\sqrt{1/n+1/n^2}\, -\, \sqrt{1/n+2/n^2}}\)

In which the limit is 0/0...it is still undefined...
Is it suppose to converge?

LOOK AT JUST THE ONE PART OF THE EXPRESSION
\(\displaystyle \sqrt{n+2}\, -\, \sqrt{n+1}\)

Then multiply both numerator and denominator of that part by
\(\displaystyle \sqrt{n+2}\) + \(\displaystyle \sqrt{n+1}\)

 

[slydez]

LOOK AT JUST THE ONE PART OF THE EXPRESSION
\(\displaystyle \sqrt{n+2}\, -\, \sqrt{n+1}\)

Then multiply both numerator and denominator of that part by
\(\displaystyle \sqrt{n+2}\) + \(\displaystyle \sqrt{n+1}\)

Oh I see. I multiplied \(\displaystyle \sqrt{n+1}\, -\, \sqrt{n+2}\) instead.

Thank you so much got the answer to \(\displaystyle 1 -\, \sqrt{2}\).

 

[lookagain]

Ohh didn't think of that.
Okay so I got 1-sqrt(2)-sqrt(n+1)+sqrt(n+2) which as n approaches 0 is infinity...

No, you would get \(\displaystyle \ -\sqrt{2} \ + \ 1. \ ** \ \ \ \)The pattern of three terms in each column adding to zero continues to the left.


Does that mean it is divergent by Telescoping Series?

If it does can the Test for Divergence still apply from my original work?

** \(\displaystyle \ \ Or, \ \ it \ \ equals \ \ \ 1 \ - \ \sqrt{2} \ \) for the sum of the series.


.