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How to prove this: $$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$$

From Apostol's number theory text i know that $$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ But how can i use this to prove my claim.

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It's not clear to me: do you want a proof using the fact from Apostol, or are you looking for any proof and you simply thought the Apostol fact would be helpful? – Matthew Conroy Dec 1 '10 at 6:40
@Matthew: Any proof would be helpful. – anonymous Dec 1 '10 at 7:54
@CHandru: I am curious. Where did you come across that lower bound? – Aryabhata Dec 2 '10 at 2:59
@Moron: Received it as an exerciese – anonymous Dec 2 '10 at 7:09
up vote 16 down vote accepted

We can use your second estimate

$$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + R(x)$$

where $\displaystyle R(x) = \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$

to show the asymptotic estimate:

$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \mathcal{\theta}\left(\frac{\sqrt{x}}{\log x}\right)$$

In fact something stronger is known, that

$\displaystyle R(x) = \mathcal{O}\left(\frac{1}{\log^2{x}}\right)$

and I believe this gives us

$$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log ^2x}\right)$$

For this, we use Abel's Identity.

Define $\displaystyle A(x) = \sum\limits_{n \leq x} \frac{a(n)}{n}$, where $\displaystyle a(n) = 1$ iff $\displaystyle n$ is prime and $\displaystyle 0$ otherwise. The above estimate tells us that $\displaystyle A(x) = \log\log x + A + R(x)$.

Setting $\displaystyle f(t) = \sqrt{t}$ and using Abel's Identity we obtain

$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} + C = A(x) \sqrt{x} - \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t$$


$$\displaystyle \int_{3}^{x} \frac{A(t)}{2\sqrt{t}} \ \text{d}t = \int_{3}^{x} \frac{\log \log t + A + R(t)}{2\sqrt{t}} \ \text{d}t$$

$$\displaystyle = C^{\prime} + A\sqrt{x} + \sqrt{x} \log \log x - \mathrm{Ei}\left(\frac{\log x}{2}\right) + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right)$$


$$\displaystyle \int \frac{\log \log x }{\sqrt{x}} = 2\sqrt{x} \log \log x - 2\mathrm{Ei}\left(\frac{\log x}{2}\right) + K$$


$$\displaystyle \int \frac{R(x)}{\sqrt{x}} = \mathcal{O}\left(\int \frac1{\log^2 x \sqrt{x}}\right) = \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$

where $\displaystyle \mathrm{Ei}(x)$ is the exponential integral.

(All gotten using Wolfram Alpha).

Since $\displaystyle \mathrm{Ei}(x) = \frac{e^x}{x} + \mathcal{O}\left(\frac{e^x}{x^2}\right)$ (again using Wolfram Alpha), we get

$$\displaystyle \sum\limits_{3 \le p \le x} \frac{1}{\sqrt{p}} = C + R(x) \sqrt{x} + \frac{2\sqrt{x}}{\log{x}} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) = \frac{2\sqrt{x}}{\log x} + \mathcal{O}\left(\frac{\sqrt{x}}{\log^2 x}\right) $$

(Hopefully, I got the computations right).

I will leave my earlier answer here (which is more relevant to the question as asked)

An elementary lower bound

Here is one elementary way (which does not use your second estimate or the prime number theorem) which proves something slightly stronger.

Using the fact the each integer is uniquely expressible as the product of a square and a square free integer, we have that

$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \prod_{p \le n} \left(1 + \dfrac1{\sqrt{p}}\right) \sum_{k=1}^{n} \dfrac1{k}$$

Now $\displaystyle e^x \gt 1 + x$, we have that

$$\displaystyle \sum_{j=1}^{n} \dfrac1{\sqrt{j}} \leq \left(\prod_{p \le n} e^{\dfrac1{\sqrt{p}}}\right)\left(\sum_{k=1}^{n} \dfrac1{k}\right)$$

We use the estimate that

$$\displaystyle \sum_{j=1}^{n} \dfrac{1}{\sqrt{j}} \ge 2\sqrt{n-1}-2$$

and that

$$\displaystyle \sum_{k=1}^{n} \dfrac1{j} \le \log (n+2)$$

to get

$$\displaystyle 2\sqrt{n-1}-2 \leq \log(n+2) \prod_{p \le n} e^{\dfrac1{\sqrt{p}}}$$

Taking logs, we see that

$$\displaystyle \log (\sqrt{n-1}-1) - \log \log(n+2) + \log 2 \le \sum_{p \le n} \dfrac1{\sqrt{p}}$$

Since, $\displaystyle \log(\sqrt{n-1}) - \log (\sqrt{n-1} -1) \to 0$ as $\displaystyle n \to \infty$,

$\displaystyle \log \log (n+2) - \log \log n \to 0 \ \text{as} \ n \to \infty$

$\displaystyle \log (n) - \log (n-1) \to 0 \ \text{as} \ n \to \infty$ and

$\displaystyle \log 2 \gt 0$, you just need to verify you inequality for finite number of $\displaystyle n$.

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+1: I really like your derivation of the lower bound. – Derek Jennings Dec 2 '10 at 8:29
Mo, I hope you don't mind; the parentheses were bothering me... ;) – J. M. Dec 2 '10 at 12:38
@Derek: Thanks, I believe something similar was used long time back to prove that sum of reciprocals are divergent. Also, there was an error (for the estimate of $\sum 1/\sqrt{j}$, I took the estimate of $\sum \sqrt{j}$ by mistake) in the proof which I have corrected. @J.M. No worries :-) In fact, thank you! – Aryabhata Dec 2 '10 at 16:12
It's the basic idea behind the proof that appeals and the minor slip made no difference to that. I have been trying to think of another elementary starting point that would produce an equally slick proof but, so far, without success. – Derek Jennings Dec 2 '10 at 18:16
@Aryabhata: Are you sure about the estimate for $\sum 1/\sqrt{j}$? There is something wrong. – xen Sep 27 '11 at 18:02

I wasn't sure that you wanted a proof that used that fact from Apostol.

One easy method not using the result you quote from Apostol is as follows: $$ \sum_{p\le x} \frac{1}{\sqrt{p}} > \sum_{p \le x} \frac{1}{\sqrt{x}} = \frac{\pi(x)}{\sqrt{x}} > c \frac{\sqrt{x}}{\log{x}} $$ where $c$ is a constant you can get from Chebyshev or Rosser and Schoenfeld, and maybe do a little computation to take care of small $x$, and you've got it. This is a much better lower bound than what you are trying to prove.

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+1: This is easy :-) – Aryabhata Dec 1 '10 at 7:33

You can try to break up the sum according to some exponential scale in $p$, and then use the previous estimate to sum up every step of the ladder, multiplying the result by $\sqrt{p_{\min}}$ where $p_{\min}$ is the lowest prime in your ladder (you should use a smarter factor since you'll be summing each $1/p$ several times, and you want the sum of the relevant factors to be $\sqrt{p_{\min}}$).

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