Alem2000
Nov4-04, 06:34 PM
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. \sum_{n=2}^{\infty}\frac{1}{nln(n)} must show if it converges or diverges soooo this is what i did
\int_{2}^{\infty}\frac{1}{nln(n)}dn integration by parts and
u=\frac{1}{ln(n)} ,
du=\frac{1}{n(ln(n))^2} ,
v=ln(n) ,
dv=\frac{1}{x}
\int_{2}^{\infty}\frac{1}{nln(n)}dn=\frac{ln(n)}{l n(n)}-\int_{2}^{+\infty}\frac{ln(n)}{nln(n)^2}dn
my logarithms cancel out and become 1 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 \frac{1}{2} now which say it converges did I do that right?
\int_{2}^{\infty}\frac{1}{nln(n)}dn integration by parts and
u=\frac{1}{ln(n)} ,
du=\frac{1}{n(ln(n))^2} ,
v=ln(n) ,
dv=\frac{1}{x}
\int_{2}^{\infty}\frac{1}{nln(n)}dn=\frac{ln(n)}{l n(n)}-\int_{2}^{+\infty}\frac{ln(n)}{nln(n)^2}dn
my logarithms cancel out and become 1 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 \frac{1}{2} now which say it converges did I do that right?