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watchmath

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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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

  2. anonymous
    • 5 years ago
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    let me know what you think

  3. watchmath
    • 5 years ago
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    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
    • 5 years ago
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    ok let me think about it

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

  6. watchmath
    • 5 years ago
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    yes :). If you are willing to contribute there I would be very glad :D. You may ask questions there though :D

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

  9. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    The are good schools in algebra. Where are you studying right now?

  11. anonymous
    • 5 years ago
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    im getting a master's in math at cleveland state. I will probably start the phd from scratch though.

  12. anonymous
    • 5 years ago
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    well nothing is popping out at me to simplify the summation. What do you think?

  13. anonymous
    • 5 years ago
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    maybe look at it is a derivative

  14. anonymous
    • 5 years ago
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    i think that would work actually

  15. anonymous
    • 5 years ago
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    actually nevermind

  16. watchmath
    • 5 years ago
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    BTW I am in algebra program too. My research is on coding theory over ring :).

  17. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so that is why you came up with that annoying ring question!

  19. watchmath
    • 5 years ago
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    welcome to the conversation satellite :D

  20. anonymous
    • 5 years ago
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    sorry to butt in. i was looking for the fibonacci thread but i cannot find it.

  21. anonymous
    • 5 years ago
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    can you so a search on these threads? i promised a reply

  22. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    so we can do it by hand as well

  24. anonymous
    • 5 years ago
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    this was the most interesting problem i've seen on this site yet

  25. anonymous
    • 5 years ago
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    where did you find this problem?

  26. watchmath
    • 5 years ago
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    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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