Is My Root Test Solution Correct?

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SUMMARY

The discussion centers on the application of the root test for determining the convergence of the series involving the term \(\frac{n}{e^n}\). The user correctly identifies the limit \(L=\lim_{n\to\infty}\sqrt[n]{\left|\frac{n}{e^n}\right|}=\frac{1}{e}\lim_{n\to\infty} n^{\frac{1}{n}}\) and seeks assistance in evaluating \(L_1=\lim_{n\to\infty} n^{\frac{1}{n}}\). By taking the natural logarithm, the limit can be expressed as \(\ln\left(L_1\right)=\lim_{n\to\infty} \frac{\ln(n)}{n}\), which converges to 0, confirming that \(L_1=1\) and thus \(L=\frac{1}{e}\). The series converges absolutely.

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Zoey93
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Hey,

I am working on this problem over the root test but I am not sure if I am doing it right. I will attach my work to this thread and I really want someone to look over my work and see if I am doing it right. By the way I didn't finish the problem because I was not sure where to go from my last step. Thank you!

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I would agree that using the root test, we need only look at:

$$L=\lim_{n\to\infty}\sqrt[n]{\left|\frac{n}{e^n}\right|}=\frac{1}{e}\lim_{n\to\infty} n^{\frac{1}{n}}$$

I would next focus on:

$$L_1=\lim_{n\to\infty} n^{\frac{1}{n}}$$

If we take the natural log of both sides, we obtain:

$$\ln\left(L_1\right)=\lim_{n\to\infty} \frac{\ln(n)}{n}$$

From this, can you now determine $L_1$, and hence $L$?
 
Converges absolutely
 

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