20200321, 03:31  #1 
Jun 2003
1,531 Posts 
Sequence
For a given integer k, the sequence is defined as :
S(0)=0 S(n)=(1S(n1))/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 
20200321, 13:18  #2  
Feb 2017
Nowhere
3^{2}×19^{2} Posts 
Quote:
S(k) = (1S(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 ? 

20200321, 15:12  #3 
Jun 2003
4,591 Posts 

20200321, 15:18  #4 
Jun 2003
5FB_{16} Posts 
For k=3
S(0)=0 S(1)=(10)/3=1/3 S(2)=(11/3)/3=2/9 S(3)=(12/9)/3=7/27 ... Hope this helps 
20200321, 15:47  #5 
Jun 2003
4,591 Posts 
The closed form of S(n) looks to be (k^n(1)^n) / ((k+1)*k^n)

20200321, 16:12  #6  
Feb 2017
Nowhere
3^{2}·19^{2} Posts 
Quote:
For n > 0, S(n) clearly is 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) 1/(k+1) for any k > 1 whether integer or not. 

20200322, 09:46  #7 
Romulan Interpreter
Jun 2011
Thailand
2×17×251 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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