Limit of a Sequence

1. Apr 16, 2010

2h2o

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$a_{n}=\sqrt[n]{4^{n}n}$$

3. 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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2. Apr 16, 2010

NeoDevin

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.

3. Apr 16, 2010

2h2o

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!