Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can anyone help me with this?

  1. Jul 8, 2015 #1
    • Member warned about posting apparent homework in a technical forum section

    • ∑ (1/n)
    • n=1
    The sum of 1/n to infinity as n =1
     
  2. jcsd
  3. Jul 8, 2015 #2
    The series is divergent.
     
  4. Jul 8, 2015 #3

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    That's called harmonic series and it is divergent as jbstemp said. But on the other hand, ##\sum_{n=1}^{\infty}\frac{1}{n^{k}}## converges if ##k## is any number larger than 1.
     
  5. Jul 8, 2015 #4
    Thanks jbstemp.
     
  6. Jul 8, 2015 #5
    Thanks hilbert2 but could you explain the question I asked in more details?
     
  7. Jul 8, 2015 #6
    Maybe tell the steps to it and does it have any awnser?
     
  8. Jul 8, 2015 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What kind of answer do you want? You have already been told the series is "divergent" which means that the infinite sum does not have any finite answer. You could prove that using the "integral test". We can extend the sum to all x rather than just integer n by using the continuous variable "x" rather than just the integer "n".
    Looking at a graph of y= 1/x, you can see that the curve, between x= n and x= n+ 1, lies below the horizontal line 1/n. That tells is that the area under the curve, the integral, is less than the sum of the area of rectangles having width 1 and height 1/n, the sum of 1/n.

    That area is given by [itex]\int_1^\infty \frac{1}{x} dx= \left[ln(x)\right]_1^\infty[/itex] but that improper integral has no answer since ln(x) has no limit as x goes to infinity. There is no finite "area under the curve" so there can be no finite area of the rectangles which would be larger.
     
  9. Jul 8, 2015 #8

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Another way of seeing divergence.
    1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+1/11+1/12+1/13+1/14+1/15+1/16)+....>1+1/2+1/2+1/2+1/2+......
     
  10. Jul 8, 2015 #9

    berkeman

    User Avatar

    Staff: Mentor

    Thread closed for Moderation...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook