- #1
Alem2000
- 117
- 0
I don't know what's 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 what's 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?
[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?