Find a simplified expression for:
1/2! + 2/3! + 3/4! + … + n/(n+1)!
My immediate reflection:
E [ n/(n+1)! ]
But, it seems that something is fishy…? There maybe a trap…
u(n) the sequence defined for any positive integer n by :
Seems to converge to 1, so if that was the series you asked, it is 1, I’m going to show it.
Looks like u(n)=[(n+1)!-1]/(n+1)!, let this relation be denoted P(n).
Proof by induction (see and learn Gary ):
u(1)=1/2, P(1) is true.
Induction step :
Suppose it is true for some given integer k.
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).
I believe you knew that the sum is not usually noted E, but capital Sigma…
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
This means it is the square of whatever the thingy thing is divided by 2
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…
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
Oh wait I see. I’m suppose to simplify n/(n+1)
(n+1)((n+1)^-1) = 1
Something like that I have no idea. I think I made it worse. ha.
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.
The Godson. The ! is the factorial symbol, you can’t just ignore it.
Once I introduced the Euler-Mascheroni constant I felt that this was no longer going to end in simplifying the expression. I gave up.
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 email@example.com?
Tks. Gary Holmes
Actually I have no idea how to type the sigma in most texts other than using the greek keyboard…
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).
This has already been posted once, but at that time, it wasnt a general expression, but till 9!. Exact same question though…
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.
focus at n/(n+1)! = [(n+1)-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)!
titiwuts. Thank you. That was very nice.
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.