Problem statement: Does the series $$\sum^\infty_{n=1} \frac{(-1)^n}{n^{1+\frac{1}{n}}}$$ diverge, converge, or converge absolutely?

EDIT: I appreciate all the answers so far. In my book, we haven't reached the Dirichlet test, Leibniz test, or derivatives, so if anyone has a line of reasoning that does not use these techniques, I would be interested to see it.

Solution attempt: We have $$\left|\frac{(-1)^n}{n^{1+\frac{1}{n}}}\right|=\frac{1}{n^{1+\frac{1}{n}}}=\frac{1}{n^{\phantom{\frac{1}{n}}}}\frac{1}{n^{\frac{1}{n}}}$$ Let ${\{b_n\}}=\frac{1}{n^{\frac{1}{n}}}$. Then $\frac{1}{\{b_n\}}=n^{\frac{1}{n}}\to 1,$ so $\frac{1}{\{b_n\}}$ is bounded, but also $\sum{|\frac{1}{n}|}$ diverges. By Theorem 26.4(ii) in Foundations of Mathematical Analysis by Johnsonbaugh, the series does not converge absolutely.

Thus, we need to determine if the series converges conditionally or diverges. At this point, I tried applying the root test, the ratio test, and the comparison test, but I didn't see anything that was tractable ... but maybe someone else does.

To apply the alternating series test, I would need to argue that $\frac{1}{n^{1+\frac{1}{n}}}$ is decreasing with limit zero. I think that this sequence does have limit zero, but I'm not sure how to show that it is decreasing, since I would need to compare $a_{n+1}=\frac{1}{(n+1)^{1+\frac{1}{n+1}}}$. I'm not sure how to handle this because $(n+1)$ is bigger, but $1+\frac{1}{n+1}$ is a smaller exponent.

Any hints?


We can apply Leibniz's test for alternating series, which is a specialization of the more general Dirichlet's test.

It is easy to see using pre-calculus tools that $\frac1{n^{1+1/n}}$ decreases monotonically for $n\ge 1$ (See Appendix) with $$\lim_{n\to \infty}\frac 1{n^{1+1/n}}=0$$ and $$\left |\sum_{n=1}^N(-1)^n\right|\le 1$$ for all $N$.

Then, Dirichlet's test guarantees that $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{1+1/n}}$ converges.


To show that $a_n=\frac{1}{n^{1+1/n}}$ decreases monotonically without invoking calculus, we show that the ratio $\frac{a_{n+1}}{a_n}$ is less than $1$. Proceeding, we have

$$\begin{align} \frac{a_n}{a_{n+1}}&=\frac{(n+1)^{\frac{n+2}{n+1}}}{n^{\frac{n+1}{n}}}\\\\ &=\left(\frac{(n+1)^{n(n+2)}}{n^{(n+1)^2}}\right)^{1/n(n+1)}\\\\ &=\left(\frac{\left(1+\frac1n\right)^{n(n+2)}}{n}\right)^{1/n(n+1)}\\\\ &\ge\left(1+\frac3n\right)^{1/n(n+1)}\dots\tag{Using Bernoulli's Inequality}\\\\ &\ge 1 \end{align}$$

as was to be shown!

  • $\begingroup$ I used this limit to show that the series does not converge absolutely, but I'm not sure how to use it in the next step. $\endgroup$ – StrugglingStudent42 Apr 14 '17 at 18:40
  • 2
    $\begingroup$ Indeed. I've edited to provide a way forward to show conditional convergence. $\endgroup$ – Mark Viola Apr 14 '17 at 18:53

If you write $ n^{1+\frac{1}{n}} = e^{\frac{n+1}{n}\log n}$, then

$$ \frac{d}{dx} \left( \frac{x+1}{x}\log x \right) = \frac{x + 1 - \log x}{x^2} > 0$$

since $\log x < x$ for any $x > 0$. This shows that $n^{1+\frac{1}{n}}$ increases and hence its reciprocal decreases in $n$. Since we already know that this converges to $0$, you can apply alternating series test to conclude the conditional convergence.

For a more systematical analysis, we may write

$$ n^{-\frac{1}{n}} = e^{-\frac{\log n}{n}} = 1 + \mathcal{O}\left(\frac{\log n}{n}\right).$$

So it follows that

$$ \frac{(-1)^n}{n^{1+\frac{1}{n}}} = \frac{(-1)^n}{n} + \mathcal{O}\left(\frac{\log n}{n^2}\right).$$

Again this is enough to conclude the convergence of the series.


Note that $n^{1+1/n}<(n+1)^{1+1/(n+1)}$ iff $n^{(n+1)^2}<(n+1)^{n(n+2)}$ iff $n<(1+1/n)^{n^2+2n}$. Now $(1+1/n)^n\to e$ as $n\to\infty$ so $(1+1/n)^{n^2+2n}$ will grow quite rapidly.


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