watchmath
  • watchmath
Interesting problem anyone? Compute \[\lim_{n\to\infty}\frac{1\cdot 1!+2\cdot 2!+\cdots+n\cdot n!}{(n+1)!}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
guess or proof?
anonymous
  • anonymous
just kidding.
watchmath
  • watchmath
guess is welcome too :D

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More answers

anonymous
  • anonymous
1
watchmath
  • watchmath
Good guess!! :D
anonymous
  • 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
  • anonymous
closed form i mean.
anonymous
  • anonymous
won't post until i prove it.
watchmath
  • watchmath
alrighty!
anonymous
  • 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
  • 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
  • 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
  • anonymous
yeah, I'm looking at that now actually ;)
anonymous
  • anonymous
actually just showing it works for r=1 will show it for the rest
watchmath
  • watchmath
Agree :).
anonymous
  • anonymous
I found it
anonymous
  • 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
  • anonymous
i think it is just \[\frac{(n+1)!-1}{(n+1)!}\]yes?
watchmath
  • watchmath
Awesome satellite! :D
anonymous
  • anonymous
proof by induction as soon as i figure it out.
anonymous
  • anonymous
My method doesn't work?
watchmath
  • watchmath
Your method works! Satellite shows that the expression is in fact equal to \(\frac{(n+1)!-1}{(n+1)!}\)
anonymous
  • anonymous
Oh I see
anonymous
  • anonymous
well actually satellite has not shown anything. i just said it.
anonymous
  • anonymous
so no credit yet that is for sure.
anonymous
  • anonymous
Fun problem though
anonymous
  • 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
  • anonymous
any hints?
watchmath
  • watchmath
Are you sure you want a hint. I am afraid it will soil the fun :).
anonymous
  • anonymous
Just factor
anonymous
  • anonymous
ok no hint. perhaps induction is not way to go?
anonymous
  • anonymous
oh lord i get the dumb guy award
watchmath
  • watchmath
ok, half hint. The summation on the top can be made into a telescoping sum.
anonymous
  • anonymous
hehe.. sometimes we can't see the forest for the trees. Happens to us all.
anonymous
  • anonymous
polpak got it. proof by induction done.
anonymous
  • anonymous
but now i have to think about the telescope.
watchmath
  • watchmath
\((n+1)!-1=(n+1)!-1!\) :D
watchmath
  • watchmath
I think a hint won't hurt now :D. \(n\cdot n!=((n+1)-1)\cdot n!\)

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