Question

Interesting problem anyone?
Compute
\[\lim_{n\to\infty}\frac{1\cdot 1!+2\cdot 2!+\cdots+n\cdot n!}{(n+1)!}\]

Answer 1

guess or proof?

Answer 2

just kidding.

Answer 3

guess is welcome too :D

Answer 4

1

Answer 5

Good guess!! :D

Answer 6

think i even know what it is, but the proof is eluding me so now i have something to think about for the afternoon.

Answer 7

closed form i mean.

Answer 8

won't post until i prove it.

Answer 9

alrighty!

Answer 10

Can we not just factor out an n! from each term in the top, and rewrite it as:
\[\lim_{n \rightarrow \infty} \frac{n\cdot n![\frac{1}{n\cdot n!} + \frac{2 \cdot 2!}{n\cdot n!} + ... + 1]}{n\cdot n!(1+\frac{1}{n})} = \lim_{n \rightarrow \infty} \frac{\frac{1}{n\cdot n!} + \frac{2 \cdot 2!}{n\cdot n!} + ... + 1}{(1+\frac{1}{n})} = \frac{1}{1} = 1\]

Answer 11

I guess it's not clear that the sum of all those n terms will go to 0 faster than the number of terms are growing but it seems reasonable to me.

Answer 12

Yes, I think it works. We just need to show more rigorously that
\[\lim_{n\to\infty}\frac{(n-r)\cdot(n-r)!}{n\cdot n!}=0\]
for all \(0<r<n\).

Answer 13

yeah, I'm looking at that now actually ;)

Answer 14

actually just showing it works for r=1 will show it for the rest

Answer 15

Agree :).

Answer 16

I found it

Answer 17

\[(n-1)! = \frac{n!}{n} \rightarrow \lim_{n\to\infty}\frac{(n-1)\cdot (n-1)!}{n\cdot n!} = \lim_{n\to\infty}\frac{(n-1)\cdot n!}{n^2 n!} = \lim_{n\to\infty}\frac{(n-1)}{n^2} = 0\]

Answer 18

i think it is just \[\frac{(n+1)!-1}{(n+1)!}\]yes?

Answer 19

Awesome satellite! :D

Answer 20

proof by induction as soon as i figure it out.

Answer 21

My method doesn't work?

Answer 22

Your method works!
Satellite shows that the expression is in fact equal to \(\frac{(n+1)!-1}{(n+1)!}\)

Answer 23

Oh I see

Answer 24

well actually satellite has not shown anything. i just said it.

Answer 25

so no credit yet that is for sure.

Answer 26

Fun problem though

Answer 27

i thought it would be simple induction. but i am getting stuck because i keep getting \[n(n+1)!+2(n+1)!\] and i need this to be \[(n+2)!\] which, if it is, is not clear to me.

Answer 28

any hints?

Answer 29

Are you sure you want a hint. I am afraid it will soil the fun :).

Answer 30

Just factor

Answer 31

ok no hint. perhaps induction is not way to go?

Answer 32

oh lord i get the dumb guy award

Answer 33

ok, half hint. The summation on the top can be made into a telescoping sum.

Answer 34

hehe.. sometimes we can't see the forest for the trees. Happens to us all.

Answer 35

polpak got it. proof by induction done.

Answer 36

but now i have to think about the telescope.

Answer 37

\((n+1)!-1=(n+1)!-1!\) :D

Answer 38

I think a hint won't hurt now :D.
\(n\cdot n!=((n+1)-1)\cdot n!\)