Question

Compute
\[
\lim_{n\to\infty}\frac{5+55+\cdots+\overbrace{55\ldots 5}^{n\text{ digits }}}{10^n}
\]

Answer 1

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

Answer 2

let me know what you think

Answer 3

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.

Answer 4

ok let me think about it

Answer 5

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

Answer 6

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

Answer 7

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?

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

maybe look at it is a derivative

Answer 14

i think that would work actually

Answer 15

actually nevermind

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

welcome to the conversation satellite :D

Answer 20

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

Answer 21

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

Answer 22

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)

Answer 23

so we can do it by hand as well

Answer 24

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

Answer 25

where did you find this problem?

Answer 26

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.