# For what primes $p$ does the series $1!+2!+3!+4!+ \cdots$ converge $p$-adically?

this is a question from a book I'm struggling with, please could you provide a clear proof

For what primes p does the series $1!+2!+3!+4!+ \cdots$ converge $p$-adically?

kind thanks

• I suggest you slow down on the number of questions you post simultaneously. Sometimes insight from an answer on one question will help you figure out another question. Furthermore, it's fairer to other users who also have questions waiting to be answered on the frontpage. – Ayman Hourieh Mar 30 '13 at 12:03

## 1 Answer

A series $\sum_{n=0}^\infty a_n$ with $a_n\in\Bbb Q$ (say) converges $p$-adically if and only if $$|a_n|_p\rightarrow0\qquad \text{as n\to\infty}$$ Now $$v_p(n!)=\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\cdots$$ is increasing and unbounded so that $$\lim_{n\to\infty}|n!|_p=\lim_{n\to\infty}p^{-v_p(n!)}=0.$$ Thus, the given series converges $p$-adically for all (finite) $p$.

• And there is no need for such precise calculations. If $n \ge p^k$, then surely $n!$ is divisible by $p^k$. That's all you need to show convergence. – GEdgar Mar 30 '13 at 12:30
• @GEdgar : true, but when precise results and calculations come with such little extra efforts it's always worth giving them, IMHO. – Andrea Mori Mar 30 '13 at 12:46
• $\sum_{n=0}^\infty a_n$ converges p-adically if the sequence of partial sums converges p-adically? can you not use a similar example as the harmonic series to find a case where $\sum_{n=0}^\infty a_n$ even though |a_n|_p\rightarrow0\qquad \text{as $n\to\infty$}? also what is $$v_p(n!)=\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\cdots$$? thanks – Mathproof P. Apr 6 '13 at 15:52
• Is it the order perhaps? could you elaborate on $$\lim_{n\to\infty}|n!|_p=\lim_{n\to\infty}p^{-v_p(n!)}=0.$$ Am I right in that you found the p-adic valuation of n!? but why does it -->0 – Mathproof P. Apr 6 '13 at 16:15
• I guess you are showing that $$v_p(n!)=\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\cdots$$ is going to infinity, but why is it unbounded? p keeps getting bigger and eventually when $p^i>n$ the floor terms will all be zero so the partial sums would all be the same i.e. it would be bounded? as n-->$\infty$ $v_p$ goes to infin.? but trying to understand why. thanks – Mathproof P. Apr 6 '13 at 16:23