Convergence and Limit of Sequence | Homework Equations Explained

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SUMMARY

The discussion focuses on the convergence of the sequence defined by a_{n}=\sqrt[n]{4^{n}n}. The correct limit of the sequence as n approaches infinity is established as 4, which was initially overlooked due to the omission of the coefficient 4 during logarithmic manipulation. The key step involves recognizing that the logarithm of the constant must be included in the limit calculation, leading to the conclusion that a_n converges to 4.

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Homework Statement



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

Homework Equations



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

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
[tex]\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}}[/tex]
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!
 

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