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 2020-03-21, 03:31 #1 Citrix     Jun 2003 22·5·79 Posts Sequence 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
2020-03-21, 13:18   #2
Dr Sardonicus

Feb 2017
Nowhere

32·509 Posts

Quote:
 Originally Posted by Citrix 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

?

2020-03-21, 15:12   #3
axn

Jun 2003

3×1,669 Posts

Quote:
 Originally Posted by Dr Sardonicus Do you mean
Quote:
 Originally Posted by Citrix For a given integer k
k is a parameter

 2020-03-21, 15:18 #4 Citrix     Jun 2003 62C16 Posts 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
 2020-03-21, 15:47 #5 axn     Jun 2003 10011100011112 Posts The closed form of S(n) looks to be (k^n-(-1)^n) / ((k+1)*k^n)
2020-03-21, 16:12   #6
Dr Sardonicus

Feb 2017
Nowhere

10001111001012 Posts

Quote:
 Originally Posted by Citrix 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.

 2020-03-22, 09:46 #7 LaurV Romulan Interpreter     Jun 2011 Thailand 13·733 Posts 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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