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I've been thinking about this one and I can't seem to find a way to solve it!

Does the series $$\sum_{n=1}^\infty \frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^n}$$ converge?

So far, the root test, the quotient limit test and the ratio test don't seem like they're useful. I suspect the answer comes from a direct comparison of another series but I can't seem to find one.

Any help would be appreciated!

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  • $\begingroup$ Pretty sure this diverges by the test for divergence. The terms tend to $1$ $\endgroup$ – Clayton May 14 '18 at 23:56
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I don't think it's that difficult. The individual terms seem to be approaching $1$ in the limit, so the series must diverge.

$$\frac{n^{n+1/n}}{(n+\frac1n)^n} = \frac{n^n\cdot n^{1/n}}{n^n\big(1+\frac1{n^2}\big)^n} = \frac{n^{1/n}}{\big(1+\frac1{n^2}\big)^n} \to 1.$$ Note that $\log (1+1/n^2)^n = n \log(1+1/n^2) \sim \frac 1n\to 0$.

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    $\begingroup$ Ted, it's even simpler ... Using Bernoulli's Inequality $$1\le \left(1+\frac1{n^2}\right)^n\le\frac1{\left(1-\frac1{n^2}\right)^n}\le \frac{1}{1-\frac1n}$$Now simply apply the squeeze theorem. No need for logarithms and no need for Taylor's Theorem here. $\endgroup$ – Mark Viola May 15 '18 at 0:03
  • $\begingroup$ Fair enough. What's simple for one person is often less simple for another. I could think of several ways to get that and was just trying to do it quickly. :) $\endgroup$ – Ted Shifrin May 15 '18 at 0:10
  • $\begingroup$ Ted, I was only supplementing your result. And for those who have yet to study calculus, wanted to present a pre-calculus way forward. And the term simpler was used in that context (i.e., use of pre-calculus tools only). $\endgroup$ – Mark Viola May 15 '18 at 0:12
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Nope, $$\lim\limits_{n\rightarrow \infty} \frac{n^{n+1/n}}{(n+1/n)^n} = 1 \neq 0.$$

If $\sum_{n=0}^{\infty} \frac{n^{n+1/n}}{(n+1/n)^n}$ coverged, we would have $\lim\limits_{n\rightarrow \infty} \frac{n^{n+1/n}}{(n+1/n)^n} = 0.$

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Notice that the $n$th term in the series is: $$\frac{n^{n+\frac{1}{n}}}{(n+\frac{1}{n})^{n}}=\frac{n^{n}n^{\frac{1}{n}}}{n^{n}(1+\frac{1}{n^{2}})^{n}}=\frac{n^{\frac{1}{n}}}{(1+\frac{1}{n^{2}})^{n}}$$ Taking limits as $n$ tends to $\infty$ results in $\frac{1}{1}=1$. By the Vanishing Condition, the series diverges.

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  • $\begingroup$ You have a typo in the last equality. :) $\endgroup$ – Ted Shifrin May 15 '18 at 0:11
  • $\begingroup$ @TedShifrin Oops, thank you! :) $\endgroup$ – Nina May 15 '18 at 16:21

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