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Brain Fart

  1. Nov 4, 2004 #1
    I dont know whats wrong with me...I have been studing for 6 hrs straight...and i noticed I have been making stupid mistakes...so can somone tell me whats up with my work...this is what I did. [tex]\sum_{n=2}^{\infty}\frac{1}{nln(n)}[/tex] must show if it converges or diverges soooo this is what i did

    [tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex] integration by parts and

    [tex] u=\frac{1}{ln(n)}[/tex] ,
    [tex]du=\frac{1}{n(ln(n))^2}[/tex] ,
    [tex]v=ln(n)[/tex] ,
    [tex]dv=\frac{1}{x}[/tex]

    [tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex]=[tex]\frac{ln(n)}{ln(n)}-\int_{2}^{+\infty}\frac{ln(n)}{nln(n)^2}dn[/tex]

    my logarithms cancel out and become [tex]1[/tex] and my integral of logorithms the top cancels and becomes one and i add it to the other side divide by two and i get [tex]\frac{1}{2}[/tex] now which say it converges did I do that right?
     
  2. jcsd
  3. Nov 4, 2004 #2
    darn it i realized what I have been doing wrong! :surprised

    bad integration i should have done [tex]u=ln(n)[/tex]
    [tex]du=\frac{1}{n}dx[/tex]
    plug and chug back into [tex]\int_{2}^{\infty}\frac{1}{nln(n)}dn[/tex] and I got[tex] \int_{2}^{\infty}\frac{1}{nu}ndu[/tex] which becomes [tex]\int_{2}^{\infty}\frac{1}{u}du[/tex] and eazy sailing after that ay?
     
    Last edited: Nov 4, 2004
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