20200321, 03:31  #1 
Jun 2003
2^{2}·5·79 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}·509 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
3×1,669 Posts 

20200321, 15:18  #4 
Jun 2003
62C_{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
1001110001111_{2} 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
1000111100101_{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
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...

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A new sequence  devarajkandadai  Miscellaneous Math  3  20201201 22:08 
Primes in nfibonacci sequence and nstep fibonacci sequence  sweety439  And now for something completely different  17  20170613 03:49 
A New Sequence  devarajkandadai  Math  3  20070320 19:43 
What's the next in the sequence?  roger  Puzzles  16  20061018 19:52 
Sequence  Citrix  Puzzles  5  20050914 23:33 