## watchmath 5 years ago Compute $\lim_{n\to\infty}\frac{5+55+\cdots+\overbrace{55\ldots 5}^{n\text{ digits }}}{10^n}$

1. anonymous

Looking at the top we can factor out a 5. 5(1+11+111+1111+...) =5*(n+10*(n-1)+100*(n-2)+...) we can write this as a sum $\sum_{i=0}^{n}10^{i}*(n-i)$ I just used maple to compute the sum for me. The result is: $n*10^{n+1}/9-(10^{n+1}/9)*(n+1)+(10/81)*10^{n+1}-n/9-10/81$ dividing out 10^n we get: 10n/9-10n/9-10/9+100/81-n/(9*10^n)-10/(81*10^n) the terms with 10^n in the denominator go to 0 and after simplifying and remembering to multiply by our factor of 5 we get: 50/81

2. anonymous

let me know what you think

3. watchmath

Yes, it seems good! Can we try to figure out the computation so we can avoid using maple. I am sure we can do it by hand.

4. anonymous

ok let me think about it

5. anonymous

side note, do you study modern algebra? I was looking at your site and saw some good stuff on there.

6. watchmath

yes :). If you are willing to contribute there I would be very glad :D. You may ask questions there though :D

7. anonymous

i am planning on getting a doctorate in algebra so maybe I will start to visit the site. Any way we can split the sum into a geometric series and one that is not quite as easy to compute. $n*\sum_{i=0}^{n}10^{i}+\sum_{i=0}^{n}-i10^{i}$ i can compute the first do you know how to compute the second?

8. watchmath

it seems we need to split the second one into several sigmas too. Do you already have some school in your mind?

9. anonymous

im looking at ohio state, university of michigan, university of illinois, university of chicago, kent, case western. i will apply to those i think. Im trying to write the second sum out to see if it can be split nicely.

10. watchmath

The are good schools in algebra. Where are you studying right now?

11. anonymous

im getting a master's in math at cleveland state. I will probably start the phd from scratch though.

12. anonymous

well nothing is popping out at me to simplify the summation. What do you think?

13. anonymous

maybe look at it is a derivative

14. anonymous

i think that would work actually

15. anonymous

actually nevermind

16. watchmath

BTW I am in algebra program too. My research is on coding theory over ring :).

17. anonymous

cool that is what my current algebra professor studies. I changed my mind again I think looking at it as a derivative we can solve by hand

18. anonymous

so that is why you came up with that annoying ring question!

19. watchmath

welcome to the conversation satellite :D

20. anonymous

sorry to butt in. i was looking for the fibonacci thread but i cannot find it.

21. anonymous

can you so a search on these threads? i promised a reply

22. anonymous

derivative of geometric series... $(n+1)a^{n}*(1/(a-1))+a^{n+1}*(-1/(a-1)^{2})+1/(a-1)^{2}$ then multiply by a. and that's the second sum (a=10)

23. anonymous

so we can do it by hand as well

24. anonymous

this was the most interesting problem i've seen on this site yet

25. anonymous

where did you find this problem?

26. watchmath

hi rsvitale you were right. Actually I did that derivative problem from some other guy yesterday. I found that $\frac{x}{(1-x)^2}=\sum_{n=1}^\infty nx^n$ and our sum is of the form $$5\sum_{i=0}^n (n-i)10^{(i-n)}=5\sum_{j=0}^n j 10^{-j}$$ As $$n\to\infty$$ we have $$5\frac{(1/10)}{(1-(1/10))^2}=\frac{50}{81}$$ which is agree to your calculation.