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This exercise specifically requires that we use the root test to determine whether the series converges or not.

All I've done so far is get the sequence in this form: $$\sqrt[n] \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^{n}} = \sqrt \frac{n^{(1+n^{-2})n}}{(n+\frac{1}{n})^n}=\frac{n^{1+n^{-2}}}{n+\frac{1}{n}} = \frac{n^{\frac{n^2+1}{n^{2}}}}{n^{2}+1}$$

But, I'm not even sure if I'm on the right track here. Any guidance is appreciated.

  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – cpiegore Mar 21 '23 at 20:31
  • I showed my effort. I didn't tell anyone to solve it for me. Not sure why the downvotes.. @cpiegore –  Mar 21 '23 at 20:34
  • You didn't provide enough context. It is good that you showed what you have already tried. That is a step in the right direction, but you should try to do more than that, if you can. In any case, you may be using the wrong test. For example, you may want to consider using the ratio test instead. – cpiegore Mar 21 '23 at 20:43
  • Right, but the exercise specifically requires the use of the root test, unfortunately. –  Mar 21 '23 at 20:46
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    OK, so do the next step. Simplify the algebra. – Ted Shifrin Mar 21 '23 at 20:50

4 Answers4

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$$\sqrt[n] \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^{n}} = \frac{n^{1 + \frac{1}{n^2}}}{n + \frac{1}{n}}=\frac{n^{1 + \frac{1}{n^2}}}{n(1 + \frac{1}{n^2})}$$

Let $x = 1 + \frac{1}{n^2}$, then

$$\frac{n^{1 + \frac{1}{n^2}}}{n(1 + \frac{1}{n^2})} = \frac{n^x}{n\cdot x}$$

See, that:

$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2}) = 1$$

So this means, that eventually, the numerator and denominator will be equal for $n$ to infinity, which means that

$$\lim_{n \rightarrow \infty} \sqrt[n] \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^{n}} = 1$$

But as $\frac{n^{1 + \frac{1}{n^2}}}{n(1 + \frac{1}{n^2})} = \frac{n^x}{nx}$ will always be greater than one ($n^x > nx$ here for $n \in \mathbb{N}$), the root test tells you that the series is divergent (as the limit "strictly approaches from above").

anon
  • 625
3

If I am not mistaken $$A=\sqrt[n] \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^{n}}= \frac{n^{2+\frac{1}{n^2}}}{n^2+1}$$ Taking logarithms and using Taylor expansion $$\log(A)=\frac{\log (n)-1}{n^2}+\frac{1}{2 n^4}+O\left(\frac{1}{n^5}\right)$$ Continuing $$A=e^{\log(A)}=1+\frac{\log (n)-1}{n^2}+O\left(\frac{1}{n^4}\right)$$

Claude Leibovici
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You are on the right track, now you just need to determine the limit of the sequence you got and see what the root test tells you about your series. hint: $\frac{n^{1+n^{-2}}}{n+\frac{1}{n}} = \frac{(\sqrt[n]{n})^{\frac{1}{n}}}{1+\frac{1}{n^2}}$

amitb03
  • 131
2

From your third step we have

$$\frac{n^{1+\frac1{n^2}}}{n+\frac{1}{n}}=\frac{n^{\frac1{n^2}}}{1+\frac{1}{n^2}} \to 1$$

indeed $n^{\frac1{n^2}}=e^{\frac{\log n}{n^2}} \to 1$.

Moreover we have that eventually

$$a_n=\frac{n^{\frac1{n^2}}}{1+\frac{1}{n^2}}=\sqrt[n^2]{\frac{n}{\left(1+\frac1{n^2}\right)^{n^2}}}>1$$

since $\left(1+\frac1{n^2}\right)^{n^2}<3$ (refer here for the proof) from which we can conclude by root test.

user
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