Convergence and Limit of Sequence | Homework Equations Explained

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Homework Statement



Show convergence or divergence, find the limit of the sequence.

Homework Equations



<br /> <br /> a_{n}=\sqrt[n]{4^{n}n}<br /> <br />

The Attempt at a Solution



See the attached picture. Sorry, it was taking way longer to figure out how to input the TeX than it was to just snap and crop a picture of my work.

My question is: where did I lose the coefficient 4? The correct answer is supposed to be convergent, a_n --> infinity = 4
 

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In the step
\lim_{x\rightarrow\infty} y = 4\lim_{x\rightarrow\infty}x^{\frac{1}{x}} \Leftrightarrow \lim_{x\rightarrow\infty}\ln y = 4\lim_{x\rightarrow\infty}\ln x^{\frac{1}{x}}
you forgot to take the logarithm of the 4 on the right hand side.
 
True, but how does that affect the outcome? The limit(x-->infinity) of 1/x = 0 thus negating anything else that multiplies against it, which still leads me back to e^0 = 1.

NvM, I see that I need to put the 4 back inside the limit then with log properties it becomes ln4+ln(1/x)... e^ln4 = 4... therefore a_n-->4. It's not always obvious to me when I should factor constants out or just leave them be.

Thanks for pointing that out, NeoDevin!
 
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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