20200321, 03:31  #1 
Jun 2003
5E5_{16} 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
5716_{8} 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
2^{2}×11×103 Posts 

20200321, 15:18  #4 
Jun 2003
1509_{10} 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
2^{2}×11×103 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
101111001110_{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
10000010111000_{2} 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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