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

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

Mathematics
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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
let me know what you think
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.

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ok let me think about it
side note, do you study modern algebra? I was looking at your site and saw some good stuff on there.
yes :). If you are willing to contribute there I would be very glad :D. You may ask questions there though :D
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?
it seems we need to split the second one into several sigmas too. Do you already have some school in your mind?
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.
The are good schools in algebra. Where are you studying right now?
im getting a master's in math at cleveland state. I will probably start the phd from scratch though.
well nothing is popping out at me to simplify the summation. What do you think?
maybe look at it is a derivative
i think that would work actually
actually nevermind
BTW I am in algebra program too. My research is on coding theory over ring :).
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
so that is why you came up with that annoying ring question!
welcome to the conversation satellite :D
sorry to butt in. i was looking for the fibonacci thread but i cannot find it.
can you so a search on these threads? i promised a reply
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)
so we can do it by hand as well
this was the most interesting problem i've seen on this site yet
where did you find this problem?
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.

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