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## 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)!

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This post is under “MathsChallenge” and has 17 respond so far.
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### 17 Responds so far- Add one»

1. 1. Guest Said：

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

2. 2. Gary Holmes Said：

My immediate reflection:

E [ n/(n+1)! ]

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

Gary

3. 3. Karys Said：

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).

4. 4. Karys Said：

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

5. 5. The Godson Said：

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

6. 6. The Godson Said：

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.

7. 7. Chris Said：

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.

8. 8. Chris Said：

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

9. 9. Karl Sharman Said：

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

10. 10. Gary Holmes Said：

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

11. 11. Karys Said：

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

12. 12. Chris Said：

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).

13. 13. Vago Said：

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

14. 14. Chris Said：

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.

15. 15. titiwuts Said：

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)!

16. 16. Chris Said：

titiwuts. Thank you. That was very nice.

17. 17. S.Javad Hashemi Said：

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.

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