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Old 2020-03-21, 03:31   #1
Citrix
 
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For a given integer k, the sequence is defined as :-

S(0)=0
S(n)=(1-S(n-1))/k

What is the formula for the nth term?
Show that for large values of n the nth term converges on 1/(k+1) for k>1

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Old 2020-03-21, 13:18   #2
Dr Sardonicus
 
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Quote:
Originally Posted by Citrix View Post
For a given integer k, the sequence is defined as :-

S(0)=0
S(n)=(1-S(n-1))/k

What is the formula for the nth term?
Show that for large values of n the nth term converges on 1/(k+1) for k>1

Do you mean

S(k) = (1-S(k -1))/k

What is the formula for the kth term?
Show that for large values of k the kth term converges on 1/(k+1) for k>1

?
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Old 2020-03-21, 15:12   #3
axn
 
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Quote:
Originally Posted by Dr Sardonicus View Post
Do you mean
Quote:
Originally Posted by Citrix View Post
For a given integer k
k is a parameter
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Old 2020-03-21, 15:18   #4
Citrix
 
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For k=3

S(0)=0
S(1)=(1-0)/3=1/3
S(2)=(1-1/3)/3=2/9
S(3)=(1-2/9)/3=7/27
...

Hope this helps
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Old 2020-03-21, 15:47   #5
axn
 
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The closed form of S(n) looks to be (k^n-(-1)^n) / ((k+1)*k^n)
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Old 2020-03-21, 16:12   #6
Dr Sardonicus
 
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Quote:
Originally Posted by Citrix View Post
For k=3

S(0)=0
S(1)=(1-0)/3=1/3
S(2)=(1-1/3)/3=2/9
S(3)=(1-2/9)/3=7/27
...

Hope this helps
Yes, thanks. I obviously misread the problem.

For n > 0, S(n) clearly is

\sum_{i=1}^n\frac{(-1)^{i-1}}{k^i}

which is a partial sum of a geometric series with first term 1/k and ratio -1/k.

Closed form for S(n) already given. S(n) \rightarrow 1/(k+1) for any k > 1 whether integer or not.
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Old 2020-03-22, 09:46   #7
LaurV
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Ha! We remember we have seen this (or similar) sometime ago in a video about a "proof" of the famous 1+2+3+...=-1/12 (let k slowly decrease to 1, to get that the limit of the sequence 1, 0, 1, 0, 1, 0,... is 0.5, practically from there start all the "layman" proofs of the above). We watched it a couple of times, and gave up after a while, something was still missing, or our brain was not developed enough...
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