Question

Silly prime question I made up cause I was bored but have the answer to, enjoy.

Answer 1

\(p_n\) is the nth prime, now I define these two functions:
\[f(n) = \frac{p_{n+1}+p_n}{p_np_{n+1}}\]
\[g(n) = \frac{p_{n+1}-p_n}{p_np_{n+1}}\]

What is:

\[\sum_{n=1}^\infty f(n)g(n)\]

Answer 2

it telescopes right?

Answer 3

Yup

Answer 4

\[\sum(\frac{1}{p_n^2}-\frac{1}{p_{n+1}^2}) \\ =\frac{1}{p_1^2}-\frac{1}{p_2^2} \\ +\frac{1}{p_2^2}-\frac{1}{p_3^2} \\ +\frac{1}{p_3^2}-\frac{1}{p^2_4} \\ - \cdots =\frac{1}{p_1^2}\]

Answer 5

where p1 is 2

Answer 6

Oh I didn't even see his response posted I was in another tab. Yeah that's exactly right. Ok so this kinda thing seems pretty simple but I guess while I got you here:

Because:
\[(u \cdot v )^2 = (u \cdot u) (v \cdot v) \cos^2 \theta\]
is saying:
\[\frac{1}{4^2} =\left( \sum_{n=1}^\infty f(n)g(n) \right)^2=\sum_{m=1}^\infty f (m)^2 \sum_{n=1}^\infty g(n)^2 \cos ^2 \theta\]

I wonder if we can find anything interesting out from this inequality:

\[\frac{1}{16} \ge \sum_{m=1}^\infty f (m)^2 \sum_{n=1}^\infty g(n)^2 \]

Answer 7

I've been sorta kicking around the more general version of the telscoping series:

\[\frac{1}{2^{2k}} = \sum_{n=1}^\infty \left(\frac{1}{p_n^k} +\frac{1}{p_{n+1}^k} \right) \left(\frac{1}{p_n^k} -\frac{1}{p_{n+1}^k} \right) \]

which leads to:
\[\frac{1}{2^{4k}} \ge \sum_{m=1}^\infty \left(\frac{1}{p_m^k} +\frac{1}{p_{m+1}^k} \right)^2 \sum_{n=1}^\infty \left(\frac{1}{p_n^k} -\frac{1}{p_{n+1}^k} \right)^2 \]

Now you might be wondering, "hey is this a legit thing to do, these are like infinite dot products yadda yadda...?" Well my answer to that is if we don't find anything interesting assuming it's true, then there's no reason to waste time making sure it's true lol

Answer 8

what qualifies as interesting :p

Answer 9

Hahaha I have no idea, but apparently primes are interesting lol

Answer 10

I guess the sum thing would be something I would try to play with
and see there is a closed form maybe...

Answer 11

IF there*

Answer 12

Right now I'm looking at each of these:

\[S_\pm = \sum_{n=1}^\infty \left(\frac{1}{p_n^k} \pm \frac{1}{p_{n+1}^k} \right)^2\]

Seems like there's some kind of interesting thing going on perhaps:


\[S_\pm = \sum_{n=1}^\infty \frac{1}{p_n^{2k}}+ \frac{1}{p_{n+1}^{2k} }\pm \frac{2}{(p_n p_{n+1})^{2k}} \]

So if you add these together:

\[\frac{1}{2}(S_++S_-) = \sum_{n=1}^\infty \frac{1}{p_n^{2k}}+ \frac{1}{p_{n+1}^{2k} } \]

Answer 13

\[\frac{1}{2} (S_++S_-) = \frac{-1}{2^{2k}} + 2 \sum_{n=1}^\infty \frac{1}{p_n^{2k}}\]

So if we let k=1 we sorta have a neat relationship to a summation of all the inverse squares of primes. Maybe we could just say that that statement is less than \(\pi^2/6\) or something and sorta combine that somehow idk, just spitballing ideas

Answer 14

lol I filled two pages up with nonsense
none of it is any good

Answer 15

Hahaha I'm sorta just filling nonsense on my papers too sadly!

Answer 16

I was trying to find a way to reorganize the following so I could get some terms to cancel:
\[f_1^2g_1^2+f_1^2g_2^2+f_1^2g_3^2+ \cdots + f_1^2 g_k^2 + \cdots + \\ f_2^2g_1^2 +f_2^2g_2^2+f_2^2g_3^2+ \cdots + f_2^2 g_k^2 + \cdots + \\ f_3^2 g_1^2+f_3^2 g_2^2+f_3^2g_3^2+ \cdots +f_3^2 g_k^2 + \cdots + \\ \cdots + \\ f_k^2 g_1^2 +f_k^2g_2^2+f_k^2g_3^2+ \cdots +f_k^2 g_k^2+ \cdots\]

but it didn't seem to work for me

Answer 17

Ahhh ok I see what you're doing, it's nice and symmetric but I couldn't get anywhere with that but I didn't pursue it very far

Answer 18

One thing I found, if my own personal notation isn't too confusing, I'll just redefine it here:

\[S_{\pm} = \sum_{n=1}^\infty \left(\frac{1}{p_n^k} \pm \frac{1}{p_{n+1}^k} \right)^2\]

From the dot product we have this:
\[\frac{1}{16^k} \ge S_+ S_-\]

But we also see from writing it that:

\[S_+>S_-\]

So multiply both sides of that by \(S_-\) to get:
\[S_+S_-> (S_{-})^2\]so combining it we have:

\[\frac{1}{16^k}> (S_{-})^2\]
or more simply,


\[\frac{1}{8^k}> S_{-}\]

So the conclusion is:

\[\frac{1}{8^k} >\sum_{n=1}^\infty \left(\frac{1}{p_n^k} -\frac{1}{p_{n+1}^k} \right)^2\]

For convenience I expanded it out:

\[\frac{1}{8^k} >\sum_{n=1}^\infty \frac{1}{p_n^{2k}}+ \frac{1}{p_{n+1}^{2k} }- \frac{2}{(p_n p_{n+1})^{2k}} \]

Hmmm