Sum of first n factorials

In my latest number theory assignment, there was a recurrence relation defined by

\displaystyle a_1=1,\ \ a_2=3, \ \ a_{n+2}=(n+3)a_{n+1}-(n+2)a_{n} \ \ \ \forall n\in\mathbb{N}

Letting b_n=a_{n}-a_{n-1} we can manipulate the recurrence relation to solve for b_n and then solve back for a_n. The result is that

\displaystyle a_n=\sum_{j=1}^n j! \ \ \ \forall n\in\mathbb{N}

So it is pretty neat fact that sum of first n factorials satisfies reasonably simple recurrence relation. Now, let’s take a look at few values of a_n.

\displaystyle a_1=1!=1

\displaystyle a_2=1!+2!=3

\displaystyle a_3=1!+2!+3!=9

\displaystyle a_4=1!+2!+3!+4!=33

\displaystyle a_5=1!+2!+3!+4!+5!=153

\displaystyle a_6=1!+2!+3!+4!+5!+6!=873

\displaystyle a_7=1!+2!+3!+4!+5!+6!+7!=5913

One might ask what interesting properties these numbers might have. Probably lots!

Observe that a_1=1 and a_3=9 are perfect squares. For what other values of n the number a_n is perfect square? It turns out that these are the only ones, and in particular a_n is not perfect square for every n\ge 4. Proving this is actually easier than it sounds. When I first saw this problem, I tried to show that a_n is strictly between two consecutive squares (which, by the way, is very useful trick in problems like this), but that didn’t go too well. There is much easier approach, as outlined below.

Let’s take a look at the last digit of a_n. It appears that a_n ends with digit 3 for n\ge 4. How can we prove this? Well, we know a_4 ends with 3, (namely a_4=33). But now observe that for j\ge 5, the term j! will end with 0, because product will contain both 2 and 5. Hence, for n\ge 5,

\displaystyle a_n=\sum_{j=1}^n j!=33+\sum_{j=5}^n j!

which therefore ends with digit 3. It is easy to check that no perfect square can end with digit 3 (namely by checking all squares mod 10). This shows that a_n is not perfect square for n\ge 4. Indeed, the solution didn’t require anything higher than basic high school math!

As a bonus, the reader of this blog can try figuring it out when the sum of first n factorials is perfect integer power.

Advertisements
This entry was posted in Uncategorized and tagged , , . Bookmark the permalink.

3 Responses to Sum of first n factorials

  1. Huseyn says:

    Thanks a lot mate. Keep it up. Best wishes from homeland, Azerbaijan.

  2. Derek O says:

    Great proof! What about a cube?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s