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Sum from n=1 to infinity of sqrt(n)/(n^2 + 1)

  1. Apr 10, 2013 #1

    Ʃ √(n)/(n2 + 1)

    Find if it converges.

    2. I'm wondering if I can rewrite this by bringing the n1/2 down to the denominator, making it negative...

    1/(n-1/2)(n2 + 1)

    = 1/(n-1 + n-1/2)

    = n + √(n)

    ...And it seems to me that this one would diverge because the n value gets higher and higher, so it would go to infinity...

    Is this the way to go about it...?

    Thank you very much! :)
  2. jcsd
  3. Apr 10, 2013 #2


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    You are doing some ghastly errors with algebra there. That's how you turned a series which does decrease with n to one that doesn't. Try and figure out what they are. And the correct way to do something like this is think about doing a comparison test first.
  4. Apr 10, 2013 #3
    ...Would you please remind me? I know that in some situations, you can bring a number with a power down to the denominator, but you have to change the power to negative.

    And then I thought that if there were already things in the denominator, then you have to multiply them by what you just brought down...

    In what situations can you do something like this? :/

    I was sure there was something in algebra at least somewhat like this...? :(
  5. Apr 10, 2013 #4


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    You can bring a factor down into the denominator as a negative power, sure. Here's where things start going wrong. What is n^(-1/2)*n^2? Is 1/(n^(-1)+n^(-1/2))=n+n^(1/2)?? Using other symbols, is 1/(1/a+1/b)=a+b???
  6. Apr 10, 2013 #5


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    Hi Lo.Lee.Ta.! :smile:

    Sorry, but you can't do a + b = 1/(1/a + 1/b). :frown:

    (and your "a" is wrong anyway :redface:)
  7. Apr 10, 2013 #6
    I tend to have difficulty choosing what my bn should be in a limit comparison test...

    In this case, it wouldn't be good to choose 1/n2 because even though it converges, it's smaller than what our original expression is!

    So that doesn't help me at all. It would only help me if bn converged but was greater than my original expression! :/

    Can't compare it to 1/√(n) either! That would diverge but is also greater than my original expression!

    UGH. How do I figure out what bn to choose? #=_=
  8. Apr 10, 2013 #7


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    Yes, you can drop the 1 because that will make it larger. Now you want to prove sqrt(n)/n^2 converges. It's a p-series. Try again to move sqrt(n) into the denominator. Correctly this time.
  9. Apr 10, 2013 #8
    Oh, for some reason I multiplied the exponents, when I should have added them! O_O

    1/((n-1/2)(n2 + 1))

    should (hopefully) equal: 1/(n1.5 + n-1/2

    = √(n)/n3/2

    Now, I guess I have to come up with a bn to compare it to...
  10. Apr 10, 2013 #9


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    No, no! You moved the sqrt(n) into the denominator. It's gone now. We were going to drop the 1 and make it larger. You just have 1/n^(3/2). Like I said that's a p-series, it has the form 1/n^p with p=3/2. You've probably learned something about them. Does it converge or diverge?

    You can't change 1/(n1.5 + n-1/2) into sqrt(n)/n^(3/2) for reasons we've already discussed.
  11. Apr 10, 2013 #10
    Alright, so I guess you can't move variables with powers unless the operation is multiplication or division, right?
    x3/x2 = 1/((x-3)(x2)) = x-1 = 1/x

    1/(x-2y3) = x2/y3

    1/(x2+y3) = y-3/x2 <--WRONG

    Okay, then with this:

    1/[(n-1/2)(n2 + 1)]

    Distributing the n-1/2...

    = 1/[n1.5 + n-1/2]

    Yes, the n-1/2 CANNOT be moved to the numerator.
    So, that's it.

    BUT HOW can we just drop the 1 in the denominator?!

    Are you already trying to find a bn or something...?

    I mean, I feel like an should equal: 1/(n1.5 + n-1/2)

    and then we might say bn = 1/n1.5

    an ≤ bn

    Since the bigger bn converges (by p-series rule), then an must also converge.

    Is this right? Thanks! :)
  12. Apr 10, 2013 #11


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    Yes, that's it. The larger series, 1/n^(3/2) converges. So 1/(n^(3/2)+n^(-1/2)) must converge. You could have dropped the 1 to begin with since sqrt(n)/(n^2+1) is less than sqrt(n)/n^2. Now just prove sqrt(n)/n^2 converges.
  13. Apr 10, 2013 #12
    Okay! <3 Thanks, Dick :D
  14. Apr 10, 2013 #13


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    x3/x2 = x1 = x
  15. Apr 11, 2013 #14
    Oh! Thanks, Mark44, for pointing that out! O_O

    x3/x2 = 1/(x-3x2) = 1/x-1 = x

    :P Silly mistake. Thanks!
  16. Apr 11, 2013 #15


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    The above is now correct, but why would you go to all that trouble? Just carry out the division to go from x3/x2 to x. (Of course, x can't be 0.)
  17. Apr 15, 2013 #16
    I was thinking about this problem again, and now I'm wondering if there's another way to do it.

    Once we rearrange it to be in this form: [itex]\frac{1}{n^{1.5} + 1/√(n)}[/itex],
    can we take the limit at n approaches infinity right now?

    lim 1/((n1.5) + 1/√(n))

    Now, I'm wondering if we can treat 1/√(n) like a variation of the harmonic series (which diverges)?
    Well, according to the p-series test, 1/√(n) diverges anyway.

    n1.5 also diverges according to the p-series test.

    So can we say it's: 1/∞ = 0

    So, the series converges?

    Is this a viable way to work the problem?
    If not, why?

    Thanks so much! :)
  18. Apr 15, 2013 #17


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    Well, no. 1/n has the form 1/∞. It converges as sequence, not as a series. Now I'm wondering if you listened to any advice you've been given. Convince me otherwise.
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