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Consider the non-negative series

$$\sum_{n = 1}^\infty e^{-n^a}, 0 < a < 1.$$

If $a = 0$, the series is divergent, and if $a \geq 1$, by root test, it is convergent. Root test doesn't give information if $0 < a < 1$. I am inclined to believe for $0 < a < 1$, it should also be convergent, but I can't prove it at the moment. Any hint?

Zhanxiong
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2 Answers2

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Note from THIS ANSWER, that the exponential function satisfies the inequality

$$e^x\ge \left(1+\frac xm\right)^m$$

for all $x\ge 0$ and for all $m\ge 0$. Then, we have

$$e^{-n^a}\le \frac{1}{\left(1+\frac {n^a}m\right)^m} \tag 1$$

Since $(1)$ holds for all $m\ge 0$, it holds for $m>\frac1a$. Choose $ma>1$.

Noting that the $m$'th term in the binomial expansion of $\left(1+\frac {n^a}m\right)^m$ is $\frac{n^{ma}}{m^m}$, we see that the series of interest converges by comparison with $\sum_{n=1}^\infty \frac{1}{n^{ma}}$ since $ma>1$.

Community
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Mark Viola
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  • Elegant answer +1 – Zhanxiong Apr 12 '16 at 02:41
  • @Solitary Much appreciative! And note that the only tools used were the limit definition of the exponential function and Bernoulli's Inequality followed by the binomial expansion and the comparison test. -Mark – Mark Viola Apr 12 '16 at 02:45
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It sufficies to show that $e^{n^{\alpha}} > n^2$ for large enough $n$, and a comparison test works. But this is true, since taking logarithms transforms this into the inequality

$$n^{\alpha} > 2 \ln n$$

which follows from, e.g., the fact that

$$\lim_{n \to \infty} \frac{\ln n}{n^{\alpha}} = 0$$

for all $\alpha > 0$.

In fact, this shows that the series is "very" convergent: $e^{n^{\alpha}}$ grows faster than any polynomial.