Strange infinite series problem using integral test.

erjkism
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Homework Statement


I need to show that

\Sigma\frac{1}{nlnnlnlnn}

from n=27 to n=10^(100,000)

is approximately equal to 8.1
 
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i did the u substitution, u =nlnn

and i get

\int\frac{1du}{u}

which just becomes [lnu]
which becomes: [ln(lnlnn)] evaluated from 27 to 10^100,000
 
is the denominator of your summation n*ln(n)*ln(ln(n))?
 
yea it is
 
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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