I was recently cleaning up my laptop when I stumbled upon a spreadsheet that I created a while ago at school when I still had an interest in math. Anyway, I remember reaching the following identity using illegal methods on conditionally convergent series.
Let $f$ be an arithmetic function defined by: $$f(n) = \sum_{d|n; \ \sqrt n\le d \le n}(-1)^d$$
Then, $$\sum_{n = 1}^\infty (-1)^n*\frac{H_n}{n} = \sum_{n = 1}^\infty \frac{f(n)}{n} = \frac{6*ln^2(2) - \pi^2}{12}$$
where $H_n$ is the $n^{th}$ harmonic number.
[EDIT] Upon further research, I am 99% certain that the latter part of the above identity is true. It seems like an interesting result, however, I have no insight on its implications.
I cannot claim this to be true and, for all I know, it may be a useless statement, anyway. However, I found it interesting that $$\text{ if n is odd and composite, then }f(n) \lt -1 \text{ (trivial)}$$$$\text{if } f(n) \gt -1 \text{, then n is even. (trivial)}$$$$\text{ if n is odd and prime or 1, then } f(n) = -1 \text{ (trivial)}$$
[EDIT] It is actually the converse of the 2nd statement that is necessary to imply the other two converses. Sorry.
Here are 3 graphs from my spreadsheet mentioned earlier. The first graph is of the difference between $\sum_{n = 1}^k \frac{f(n)}{n} $ and $\frac{6*ln^2(2)-\pi^2}{12}$ for k up to 14000. The second and third graphs are of $f(n)$ for n up to 50 and 14000 respectively.(I do not have the reputation to post images, sorry!)
In these graphs it appears that $f(n)$ is random and growing very slowly in magnitude.
In conclusion, my questions are:
- Can anyone see a way to (dis)prove the converse of the statement about $f(n)$ for even n?
- Does anyone have references where I can find information about this from a more qualified standpoint?
- Is this all just mathematical jibberish that I wasted my time on back in the day, or is there a reason to find it interesting that the identity appears true?
Thanks.
The first equality is certainly true (modulo some convergence justification): \begin{align*} \sum_{n=1}^\infty \frac{f(n)}n &= \sum_{n=1}^\infty \frac1n \sum_{\substack{d\mid n \\ \sqrt n\le d \le n}} (-1)^d \\ &= \sum_{d=1}^\infty (-1)^d \sum_{\substack{d\le n \le d^2 \\ d\mid n}} \frac1n \\ &= \sum_{d=1}^\infty (-1)^d \sum_{1\le m \le d} \frac1{md} \\ &= \sum_{d=1}^\infty \frac{(-1)^d}d \sum_{1\le m \le d} \frac1m = \sum_{d=1}^\infty \frac{(-1)^d}d H_d. \end{align*}
If $n$ is even, then $$ f(n) = \#\{ d\mid n,\, d\text{ even},\, \sqrt n\le d \le n \} - \#\{ c\mid n,\, c\text{ odd},\, \sqrt n\le c \le n \}. $$ Moreover, every divisor $c$ of $n$ counted in the second set has a corresponding divisor $2c$ of $n$ counted in the first set; therefore the difference is definitely nonnegative, proving the desired converse. (The difference can certainly be positive: even divisors of $n$ between $\sqrt n$ and $2\sqrt n$ contribute to the first set but not the second, as do divisors of the form $4c$, $8c$, etc.)
As for the growth rate: when $n$ is odd (and, say, not a perfect square), then $f(n) = -\frac12d(n)$, where $d(n)$ is the number of divisors of $n$. And the maximal order of the divisor function is known. Something similar will be true for positive values.