Why does $\sum \frac{1}{n^{1 + \epsilon}}$ converge?

Question

The proof that the infinite sum of $\frac{1}{n}$ diverges seems to have a fair amount of breathing room. We group successive terms in quantities of increasing powers, starting with $\frac{1}{2}$, then $\frac{1}{3} + \frac{1}{4}$, then the next four terms, then the next eight terms, etc., and note that each of the groups is greater than or equal to $\frac{1}{2}$, and adding $\frac{1}{2}$ forever approaches $\infty$.

As extra credit for this proof, each group after the first is strictly greater than $\frac{1}{2}$, so the divergence actually occurs faster. Furthermore, we didn't even need the terms to be that large; adding $\frac{1}{1,000,000}$ forever would also approach $\infty$. Why then, given this generous cushion in the proof, is it the case that $\frac{1}{n^{1 + ε}}$ for some tiny ε converges? Why is the power of $n$ so fragile to nudges in the positive direction given how firmly $\frac{1}{n}$ seemed to diverge?

Answer 1

To address the question why a small $\epsilon$ is enough, note that analogously to the proof of divergence for the harmonic series, we can say that

$$\frac1{2^{1+\epsilon}} \ge \frac1{2^{1+\epsilon}}, $$

$$\frac1{3^{1+\epsilon}} + \frac1{4^{1+\epsilon}} \ge 2\frac1{4^{1+\epsilon}} = \frac1{2^{1+2\epsilon}}, $$

$$\frac1{5^{1+\epsilon}} + \frac1{6^{1+\epsilon}} + \frac1{7^{1+\epsilon}} + \frac1{8^{1+\epsilon}}\ge 4\frac1{8^{1+\epsilon}} = \frac1{2^{1+3\epsilon}}, $$

and so on. Note that the terms on the right hand side are no longer are constant as they used to be for the harmonic series, instead they form a geometric progression with the factor $\frac1{2^\epsilon}$. When $\epsilon$ is small, that value is just a tiny bit smaller than $1$.

Still any geometric sequence with factor $< 1$ will converge to $0$, even if slowly. That means if we take the sum of the right hand sides, it is no longer an infinite sum of $\frac12$ that diverges, but a geometric series that does converge!

So the main "problem" in translating the proof for $\epsilon>0$ is that our divering minorante for the harmonic series no longer divererges!

In essence it is just the difference that $\sum_{i=0}^{\infty}\frac12$ diverges, while $\sum_{i=0}^{\infty}\frac1{2^{1+i\epsilon}}$ converges.

This insight allows you to actually prove that $\sum_{n=0}^{\infty}\frac1{n^{1+\epsilon}}$ converges without the integral methods mentioned in other answers.

That's because

$$\frac1{3^{1+\epsilon}} + \frac1{4^{1+\epsilon}} \le 2\frac1{2^{1+\epsilon}} = \frac1{2^{\epsilon}}, $$

$$\frac1{5^{1+\epsilon}} + \frac1{6^{1+\epsilon}} + \frac1{7^{1+\epsilon}} + \frac1{8^{1+\epsilon}}\le 4\frac1{4^{1+\epsilon}} = \frac1{2^{2\epsilon}}, $$

$$\frac1{9^{1+\epsilon}} + \frac1{10^{1+\epsilon}} + \frac1{11^{1+\epsilon}} + \frac1{12^{1+\epsilon}} +\frac1{13^{1+\epsilon}} + \frac1{14^{1+\epsilon}} + \frac1{15^{1+\epsilon}} + \frac1{16^{1+\epsilon}} \le 8\frac1{8^{1+\epsilon}} = \frac1{2^{3\epsilon}}, $$ a.s.o.

Now our series has a converging majorante $\sum_{i=0}^{\infty}\frac1{2^{i\epsilon}}$, so converges itself.

Answer 2

I will address the last paragraph of your post. As long as epsilon is a positive fixed quantity, the series will converge. This can be seen with the Integral test. $1/x$ integrates to $lnx$ and with $x$ going to infinity, the integral and thus the series diverges. But if the exponent is more than $1$, the polynomial term integrates to another polynomial term. I leave it up to you the figure out why that implies convergence because this essentially will answer your last part of the question. Lastly (and not unimportant!) a note on the word "fixed". If the epsilon is not a fixed positive quantity, but variable, then the series may be divergent. For example the series $\frac{1}{n^{(1+1/n)}}$ has a "variable" exponent, but the exponent is more than $1$. Yet this series turns out to be divergent.

Answer 3

Also, the integral test can be used to show that $\sum \dfrac1{n\ln \ln ... \ln(n)} $ diverges for any fixed number of nexted $\ln$. This is because, if we define $\ln_0(n) = 1 $ and $\ln_{k+1}(n) =\ln(\ln_k(n)) $, then $(\ln_k(x))' =\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)} $.

Proof.

$(\ln_1(x))' =(\ln(x))' =\dfrac1{x} $ and $(\ln_2(x))' =(\ln(\ln(x)))' =(\ln(x))'\dfrac1{\ln(x)} =\dfrac1{x\ln(x)} $

If $(\ln_k(x))' =\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)} $, then

$\begin{array}\\ (\ln_{k+1}(x))' &=(\ln(\ln_k(x)))'\\ &=(\ln_k(x))'\dfrac1{\ln_k(x)}\\ &=\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)\ln_k(x)}\\ &=\dfrac1{x\prod_{j=1}^{k}\ln_{j}(x)}\\ \end{array} $

Since $=\ln_k(x) \to \infty$ as $x \to \infty$ for any fixed $k$, $\int \dfrac{dx}{x\prod_{j=1}^{k-1}\ln_{j}(x)} =\ln_k(x) \to \infty$ as $x \to \infty$ so $\sum \dfrac1{n\prod_{j=1}^{k-1}\ln_{j}(n)} $ diverges by the integral test.

You can similarly show that $\sum \dfrac1{n\prod_{j=1}^{k-1}\ln_{j}(n)\ln_{k}^{1+\epsilon}(n)} $ converges for any fixed $k$ and $\epsilon > 0$.