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

- watchmath

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- schrodinger

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- anonymous

guess or proof?

- anonymous

just kidding.

- watchmath

guess is welcome too :D

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## More answers

- anonymous

1

- watchmath

Good guess!! :D

- anonymous

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

- anonymous

closed form i mean.

- anonymous

won't post until i prove it.

- watchmath

alrighty!

- anonymous

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

- anonymous

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.

- watchmath

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

- anonymous

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

- anonymous

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

- watchmath

Agree :).

- anonymous

I found it

- anonymous

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

- anonymous

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

- watchmath

Awesome satellite! :D

- anonymous

proof by induction as soon as i figure it out.

- anonymous

My method doesn't work?

- watchmath

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

- anonymous

Oh I see

- anonymous

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

- anonymous

so no credit yet that is for sure.

- anonymous

Fun problem though

- anonymous

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.

- anonymous

any hints?

- watchmath

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

- anonymous

Just factor

- anonymous

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

- anonymous

oh lord i get the dumb guy award

- watchmath

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

- anonymous

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

- anonymous

polpak got it. proof by induction done.

- anonymous

but now i have to think about the telescope.

- watchmath

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

- watchmath

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

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