# Brain Fart

1. Nov 4, 2004

### Alem2000

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

$$u=\frac{1}{ln(n)}$$ ,
$$du=\frac{1}{n(ln(n))^2}$$ ,
$$v=ln(n)$$ ,
$$dv=\frac{1}{x}$$

$$\int_{2}^{\infty}\frac{1}{nln(n)}dn$$=$$\frac{ln(n)}{ln(n)}-\int_{2}^{+\infty}\frac{ln(n)}{nln(n)^2}dn$$

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?

2. Nov 4, 2004

### Alem2000

darn it i realized what I have been doing wrong! :surprised

bad integration i should have done $$u=ln(n)$$
$$du=\frac{1}{n}dx$$
plug and chug back into $$\int_{2}^{\infty}\frac{1}{nln(n)}dn$$ and I got$$\int_{2}^{\infty}\frac{1}{nu}ndu$$ which becomes $$\int_{2}^{\infty}\frac{1}{u}du$$ and eazy sailing after that ay?

Last edited: Nov 4, 2004