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Given the series: $\sum_{n=0}^{\infty}(-1)^{n}(\sqrt[n]{n} - 1)^{n}$. Does the series converge?

Attempt to solution (might be incorrect):

$(\sqrt[n]{n} - 1)^{n}> (1+\frac{1}{n})^{n}$

$(1+\frac{1}{n})^{n} \to e \Rightarrow (\sqrt[n]{n} - 1)^{n}$ lower-bounded by $e$. Based on Leibniz Criterion the sequence $\{A_n\}$ (in our case, $(\sqrt[n]{n} - 1)^{n}$) is monotone decreasing, but its limit is not $0$ at infinite $\Rightarrow$ series diverge.

Is it enough to say that since the sequence is lower-bounded, the limit of it at infinite is not $0$, or should I actually calculate the limit of the sequence?

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The limit of $(-1)^n (n^{1/n} - 1)^n$ goes to $0$. – Jonas Teuwen Nov 27 '10 at 22:09
To see that your inequality is false, note that it is equivalent to $\root n\of n -1>1+(1/n)$, or $n>(2+(1/n))^n$. But $(2+(1/n))^n>2^n>n$. – Andrés E. Caicedo Nov 27 '10 at 22:36

The alternatingness of the series is something of a red herring as the series converges absolutely. By the root test, to show this it suffices to show that $\lim_{n \rightarrow \infty} |n^{1 \over n} - 1| = 0$. In other words, it suffices to show that $\lim_{n \rightarrow \infty} n^{1 \over n} = 1$.

There are a few ways to show this limit is in fact $1$. One way is to note that $\ln (n^{1 \over n}) = {\ln(n) \over n}$, and the latter is seen to go to zero as $n$ goes to infinity using L'Hopital's rule. Since the natural log of the $n$th term of the sequence goes to zero, the $n$th term of the sequence goes to $e^0 = 1$.

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Your lower bound for $(n^{1/n} -1)^n$ is not correct.

Hint: To show that $(n^{1/n} -1)^n \rightarrow 0$ write $n^{1/n} = 1 + r_n$ and use the binomial theorem to show that for $n \ge 2$ we have $r_n < \sqrt{2/(n-1)}.$

Hint2: When you expand $(1+r_n)^n$ the important term in the expansion is the one in ${r_n}^2 .$

In this way you can show that the series converges absolutely, you don't really need the Leibniz criterion although they will do the job.

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Write your series as $\sum_{n = 1}^\infty (-1)^n a_n$ where $a_n = (n^{1/n} - 1)^n$. Note that $(a_n)$ is monotonically decreasing to zero and all terms are positive, hence by the alternating series test (or Leibniz' test) the series converges.

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