My goal is to prove that:

$\displaystyle\sum\limits_{n=1}^\infty \frac{1}{n^{x}} \rightarrow \infty$ when $ x\rightarrow 1^+$

My first approach (which failed) is here:

To prove $f(x)\rightarrow \infty$ with a "home made" strategy

I think the confusing part is the "$x\rightarrow 1^+$". I have now argued for the statement, but I haven't used the direction to $1$.

I begin with $s_{N}(x) = \displaystyle\sum\limits_{n=1}^N \frac{1}{n^{x}}$ and write $s_{2^{n}}(x)$ as:

$\begin{equation*} \begin{split} s_{2^{n}}(x) &= 1 \\ & + \frac{1}{2^{x}}\\ & + \frac{1}{3^{x}} + \frac{1}{4^{x}} \\ \\ \\ & \dots \\ & + \frac{1}{(2^{nx-x}+1)^x} + \frac{1}{(2^{xn-x}+2)^x}+ \ldots + \frac{1}{2^{nx}} \end{split} \end{equation*}$

From this I can deduce that there are $n+1$ lines and every line has a number of terms of the form $2^{k-1}$ and ends with a term on the form $\frac{1}{2^{kx}}$ (where all the other terms are greater than this, because $x \rightarrow 1^+$). The sum of a line (including the first line) is therefore greater than or equal $2^{k-1}\frac{1}{2^{kx}}$ which is equal to $\frac{1}{2}$ when $x\rightarrow 1^+$ (the first line is $1$)

Finally I get:

$\frac{1}{2} \cdot (n+1) \leq s_{2^{n}}(x)$

And when $n\rightarrow \infty$ then $s_{2^{n}}(x)\rightarrow \infty$

And I'm done.


Well. My suspicion is that I have forgotten something, because I havn't used the direction $x\rightarrow 1^+$. Is this a valid proof?

  • $\begingroup$ Don't you want $+$ instead of $=$ in the breakdown of $s_{2^n}(x)$? And in the last line a typical denominator should be $(2^{nx-x}+i)^x$. $\endgroup$ – Brian M. Scott May 13 '12 at 1:09
  • $\begingroup$ Oops! Yes I wanted +! I have changed it now. $\endgroup$ – bemyguest May 13 '12 at 1:12
  • $\begingroup$ I went ahead and made the other correction that I mentioned in my first comment. $\endgroup$ – Brian M. Scott May 13 '12 at 5:19
  • $\begingroup$ Just in case: have you already seen this? $\endgroup$ – J. M. is a poor mathematician May 13 '12 at 5:24

The short proof given by chessmath is perhaps the easiest way to go, but here are some comments on the approach that you’ve taken. (And congratulations on being aware of Oresme’s accomplishments!)

You can’t say that $$2^{k-1}\frac1{2^{kx}}=\frac12$$ when $x\to 1^+$; you’re talking about a limit, so you can only say that $$\lim_{x\to 1^+}2^{k-1}\frac1{2^{kx}}=\frac12\;.$$ This is of course true no matter how $x\to1$, but it doesn't justify the rest of your argument. We already know that the series diverges when $x=1$ and converges for $x>1$; the problem is to show that its sum increases without bound as $x\to1^+$.

For this you can adapt your argument. You’ve shown that $$s_{2^n}(x)\ge\sum_{k=0}^n\frac{2^{k-1}}{2^{kx}}\;;$$ now just extend this reasoning to argue that $$s(x)=\sum_{n=1}^\infty\frac1{n^x}\ge\sum_{k\ge 0}\frac{2^{k-1}}{2^{kx}}=\frac12\sum_{k\ge 0}\left(\frac2{2^x}\right)^k=\frac12\sum_{k\ge 0}\left(\frac1{2^{x-1}}\right)^k\;.\tag{1}$$

Here is where you use the fact that $x\to1^+$: $x>1$, so $\dfrac1{2^{x-1}}<1$, and the geometric series in $(1)$ converges, giving $$s(x)=\sum_{n=1}^\infty\frac1{n^x}\ge\frac12\left(\frac1{1-1/2^{x-1}}\right)=\frac12\left(\frac{2^{x-1}}{2^{x-1}-1}\right)\ge\frac12\left(\frac1{2^{x-1}-1}\right)\;,\tag{2}$$ since $2^{x-1}>1$ for $x>1$. Now simply observe that as $x\to1^+$, $2^{x-1}-1\to0^+$, and the lower bound in $(2)$ increases without bound, so that we must have $\lim\limits_{x\to1^+}s(x)=\infty$.

  • $\begingroup$ Thank you so much! That's clever! I will be more careful next time I use the limit defination. $\endgroup$ – bemyguest May 13 '12 at 9:15

I would prove that by definition

Let given an $M>0$ there is and N>0 such that $\sum _{n=1} ^N\frac{1}{n}>2M$. Since the $f(x)=\sum _{n=1} ^N\frac{1}{n^x}$ function is continuous in $x$ there is a $\delta>0$ such that $1<x<1+\delta$ implies $f(x)>M$ then

$$\sum _{n=1} ^\infty\frac{1}{n^x}>f(x)>M.$$

PS: For your approach you should use that $\sum _{n=1} ^\infty\frac{1}{n^x}\geq \sum_{k=1}^{\infty}(2^{1-x})^k=\frac{2^{1-x}}{1-2^{1-x}}\to \infty$ when $x\to 1^+$.

Observe that the sum is given by that expression only when $x>1$.

  • $\begingroup$ Is there any particular reason why you put $\sum_{n=1}^N \frac 1n>2M$ instead of just $M$? $\endgroup$ – Yongyi Chen May 13 '12 at 3:09
  • $\begingroup$ No,just a mania because to prove that we use the definition more directly. $\endgroup$ – checkmath May 13 '12 at 4:28

The following proof is from the general theory of Riemann's Zeta Function:

1) Put $\,\,\displaystyle{\zeta(x):=\sum_{n=1}^\infty\frac{1}{n^x}\,\,,\,\,x>1}$

2) Use the fact that the function $\,\,\displaystyle{f(y):=\frac{1}{y^x}\,\,,\,\,y>0\,\,}$ is monotone decreasing together with the integral mean value theorem to show that $\,\,\displaystyle{\frac{1}{(n+1)^x}\leq\int_n^{n+1}\frac{1}{y^x}dy\leq\frac{1}{n^x}\,\,,\,\,n\in\mathbb{N}}$

3) Add the above equalites over the naturals (check the telescopic series!) and get that $$\zeta(x)-1\leq \frac{1}{x-1}\leq\zeta(x)$$

4) Get from the above that $\,\,\displaystyle{\lim_{x\to 1^+}(x-1)\zeta(x)=1}$

5) Deduce your result.



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