What did i do wrong? Ratio and comparison test question?

rickyram
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Homework Statement



The summation of n/lnn from n=1 to infinity

Homework Equations



Ratio and Comparison test


The Attempt at a Solution



http://i55.tinypic.com/2jxuue.jpg

Alright so the way labeled "right way" on the picture is the right way according to the book but isn't the way I did with the ratio test right also? What did i do wrong?
 
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It is NOT true that

\frac{ln(n+1)}{ln(n)}=ln(1)
 
ln (n+1) / ln (n) isn't ln(1), in the limit or otherwise.
 
Oh man, I got to brush up on my logarithm rules... Anyway thanks for answerin my question, it has been a big help.
 
Apply the divergence test directly. take limit n->(infinity) n/lnn. I doubt that goes to zero!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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