## Simplify the factorial series

Posted by Chris on June 14, 2010 – 11:05 am

Find a simplified expression for:

1/2! + 2/3! + 3/4! + … + n/(n+1)!

Posted by Chris on June 14, 2010 – 11:05 am

Find a simplified expression for:

1/2! + 2/3! + 3/4! + … + n/(n+1)!

June 14th, 2010 at 2:28 pm

((n+1)!-1)/(n+1)!

June 14th, 2010 at 3:16 pm

My immediate reflection:

E [ n/(n+1)! ]

But, it seems that something is fishy…? There maybe a trap…

Gary

June 14th, 2010 at 3:49 pm

Excel ———

u(n) the sequence defined for any positive integer n by :

u(1)=1/2

u(n+1)=u(n)+(n+1)/(n+2)!

Seems to converge to 1, so if that was the series you asked, it is 1, I’m going to show it.

u(2)=5/6

u(3)=23/4

u(4)=119/120

Looks like u(n)=[(n+1)!-1]/(n+1)!, let this relation be denoted P(n).

Proof by induction (see and learn Gary ):

Basis :

u(1)=1/2, P(1) is true.

Induction step :

Suppose it is true for some given integer k.

u(k+1)=[...]/(k+1)!+(k+1)/(k+2)!

u(k+1)=[(k+2)!-(k+2)+(n+1)]/(k+2)!

u(k+1)=[(k+2)!-1]/(k+2)!

P(k+1) is true if P(k) is.

Therefore P(n) is true for any given positive integer n.

We then easily show that it converges :

u(n)=1 – 1/(k+2)!

1/(k+2)! converges to 0.

1 is 1. (duh)

u(n) converges to 1.

Gary, the sum notation is not a real simplification. It uses much less ink/your keyboard, but it is the same as the given developped form. You could think that way : If you asked someone or a computer to calculate it, what would you ask it to do ?

With or without the sum notation, you would anyway ask to have the SUM of all terms i/(i+1)! with i increasing from 1 to n. So that’s the same. Simplification means simpler calculus. Sum = you must go through all u(k) until u(n), and now a simple division gives you any u(n).

June 14th, 2010 at 3:50 pm

I believe you knew that the sum is not usually noted E, but capital Sigma…

June 14th, 2010 at 6:59 pm

1/2 + 2/3 + 3/4 Is sort of like half a triangle like below without the potential square:

1/2 + 1/3 + 1/4

+ 1/3 + 1/4

+ 1/4

This means it is the square of whatever the thingy thing is divided by 2

hense…

(n/2)(n+1)

The plus one is there beacuse of the extra slice of the triangle

oh yeah, the original number starts at half and is going down so therefore…

n/(n+1)

oh wait thats the thingy already written up at the top. Oh well. Wait what? thats the answer. I’m so lost I don’t even no what the question is anymore n(n+1) is the answer to the question yet it is in the question. Perhaps I’ve overlooked something. After all I am tired and sick of typing. Yeah I’m done here. THE END goodnight

June 14th, 2010 at 7:27 pm

Oh wait I see. I’m suppose to simplify n/(n+1)

(n+1)((n+1)^-1) = 1

Therefore…

n((n+1)^-1)

n(1/(n+1))

Something like that I have no idea. I think I made it worse. ha.

June 14th, 2010 at 7:29 pm

Karys, you’ve done it. The simplified form is:

u(n) = [(n+1)!-1]/(n+1)! = 1 – 1/(n+1)!

Small point; you mentioned calculus, even though you haven’t used it.

June 14th, 2010 at 7:40 pm

The Godson. The ! is the factorial symbol, you can’t just ignore it.

June 15th, 2010 at 5:58 am

Once I introduced the Euler-Mascheroni constant I felt that this was no longer going to end in simplifying the expression. I gave up.

Karl

June 15th, 2010 at 1:14 pm

1) This is in response to #3 Karys:

U r a nice teacher. And patient too, in addition to the fact that u seem 2b an intelligent person… And picky too.

2) Response to #4 Karys:

How to type “Greek Sigma” (18th Grk Ltr) on English screen, pls?

As u know: I typed “E” only cuz it looks like Sigma, although E is supposed 2b Epsilon (5th Grk Ltr)…

3) I’m cheating, by copying from Dictionary: “Σ”

4) Karl & Karys,

I like both of u. Pls kindly send me an eM to garyfan@hotmail.com?

Tks. Gary Holmes

June 15th, 2010 at 4:05 pm

Actually I have no idea how to type the sigma in most texts other than using the greek keyboard…

June 15th, 2010 at 6:14 pm

Hi Karys. In the Windows charmap applet, ∑ is U+2211. Also the ‘No-Break Space’ character is U+00A0; this is useful for making tables and indented text on this site (which infuriatingly squeezes out initial and multiple standard space characters – ruining tables).

June 15th, 2010 at 6:39 pm

This has already been posted once, but at that time, it wasnt a general expression, but till 9!. Exact same question though…

June 15th, 2010 at 7:00 pm

Hi Vago. I hadn’t seen the previous vrsion. I found it on a newspaper’s puzzle page.

I didn’t know how to deduce the result, even though I had found out (via http://www.wolframalpha.com) what it was.

I was quite impressed by Karys’s efforts and wanted to kick myself for not calculating the first few sums by hand for myself.

June 16th, 2010 at 12:21 am

focus at n/(n+1)! = [(n+1)-1]/(n+1)!

= [(1/n!)-(1/(n+1)!)]

take SUM, it’s gonna be like

S[n/(n+1)!]= (1/1! – 1/2!) + (1/2! – 1/3!) + …. + (1/(n)! – 1/(n+1)!)

= 1/1! – 1/(n+1)!

June 17th, 2010 at 3:47 pm

titiwuts. Thank you. That was very nice.

March 25th, 2012 at 12:32 am

1/2! + 2/3! + 3/4! + … + n/(n+1)!=sum((k+1-1)/(k+1)!)=

sum((k+1)/(k+1)!-1/(k+1)!)=sum(1/k!-1/(k+1)!)=1-1/(n+1)! according to Telescopic rule in series.