 mersenneforum.org Sequence
 Register FAQ Search Today's Posts Mark Forums Read 2020-03-21, 03:31 #1 Citrix   Jun 2003 3·503 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

57168 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

22×11×103 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 3×503 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 10001101101002 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

1011110011102 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

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.   2020-03-22, 09:46 #7 LaurV Romulan Interpreter   Jun 2011 Thailand 23·3·349 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 Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post sweety439 And now for something completely different 17 2017-06-13 03:49 devarajkandadai Math 1 2007-08-25 15:23 devarajkandadai Math 3 2007-03-20 19:43 roger Puzzles 16 2006-10-18 19:52 Citrix Puzzles 5 2005-09-14 23:33

All times are UTC. The time now is 05:08.

Mon Mar 30 05:08:41 UTC 2020 up 5 days, 2:41, 2 users, load averages: 1.35, 1.45, 1.44