If I am asked for the first five terms of the sequence of the partial sums for the above series, should I consider the first term with n=11? Or is the first term still with n=1? I'm thinking the first term would be with n=11. Is the first term of partial sums then (12)(3^11)/(10)(4^11) ?
I'm not sure how to find the limit. I tried the ratio test b/c its the only one I can get somewhere with. I didn't use integral test b/c getting integral of that expression looks tough. I'm shaky on how to find limits like these. However, I did use ratio test and got (3)(n+2)(n-1)/(n)(n+1)(4). I took limit and had to use L'hopitals rule a couple of times and came up with 3/4 as the limit which should mean that it converges, but to what? Please let me know if I'm on the right track. I'm confused about how to get limit and also by the fact that it starts at 11 instead of one.
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Verified answer
I can't quite figure out why the 11, but here is a way to get the limit:
first we know that for all -1<x<1
1/(1-x) = sum x^n (the sum is from 0 to ∞)
and by integrating this (since we can prove that this is legitimate)
ln(1-x) = sum x^(n+1)/(n+1)
Now
S = sum[(n+1)(3^n)/(n-1)(4^n),11, ∞]
= sum[(k+3)/(k+1) (3/4)^(k+2), 9, ∞] (k = n-1)
= 3/4 sum[(k+3)/(k+1) (3/4)^(k+1), 9, ∞]
= 3/4 sum[(1 + 2/(k+1)) (3/4)^(k+1), 9, ∞]
= 3/2 sum[(3/4)^(k+1)/(k+1), 9, ∞] + (3/4)^11 sum[(3/4)^k, 0, ∞]
= 3/2 [ln(1-3/4) - sum[(3/4)^(k+1)/(k+1), 0, 8]] + (3/4)^11/(1-3/4)
= 3^11/4^10 - 3/2 [ ln(4) + sum[(3/4)^(k+1)/(k+1), 0, 8]]
The result is not fully simplified but it would not pose any theoretical difficulties to do so.
It appears we should find something in the form
S = a + b ln(4)
where a and b are rationals.
Yes, the first term is n = 11. I've been trying this for a few hours, incidentally, but I haven't really succeeded in coming up with a definite answer. Will have to get back to you on this after I ask my lecturer for some help... I'm rather shaky myself. This might not really help, but did you try finding it via subtracting the series with n = 1 to 11 from the series with n = 1 to ∞?
Arrgh. *tries somemore*
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I can't speak Chinese. I am so bad at math it is deplorable, I need to take off my shoes to count to 11.
do you know what the answer is if you do tell me plz.