## watchmath 5 years ago Interesting problem anyone? Compute $\lim_{n\to\infty}\frac{1\cdot 1!+2\cdot 2!+\cdots+n\cdot n!}{(n+1)!}$

1. anonymous

guess or proof?

2. anonymous

just kidding.

3. watchmath

guess is welcome too :D

4. anonymous

1

5. watchmath

Good guess!! :D

6. 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.

7. anonymous

closed form i mean.

8. anonymous

won't post until i prove it.

9. watchmath

alrighty!

10. anonymous

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

11. 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.

12. 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 $$0<r<n$$.

13. anonymous

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

14. anonymous

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

15. watchmath

Agree :).

16. anonymous

I found it

17. anonymous

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

18. anonymous

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

19. watchmath

Awesome satellite! :D

20. anonymous

proof by induction as soon as i figure it out.

21. anonymous

My method doesn't work?

22. watchmath

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

23. anonymous

Oh I see

24. anonymous

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

25. anonymous

so no credit yet that is for sure.

26. anonymous

Fun problem though

27. 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.

28. anonymous

any hints?

29. watchmath

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

30. anonymous

Just factor

31. anonymous

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

32. anonymous

oh lord i get the dumb guy award

33. watchmath

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

34. anonymous

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

35. anonymous

polpak got it. proof by induction done.

36. anonymous

but now i have to think about the telescope.

37. watchmath

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

38. watchmath

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