Sequence

Citrix · 2020-03-21, 03:31

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

Dr Sardonicus · 2020-03-21, 13:18

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

?

axn · 2020-03-21, 15:12

Quote:

Originally Posted by Dr Sardonicus

Do you mean

Quote:

Originally Posted by Citrix

For a given integer k

k is a parameter

Citrix · 2020-03-21, 15:18

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

axn · 2020-03-21, 15:47

The closed form of S(n) looks to be (k^n-(-1)^n) / ((k+1)*k^n)

Dr Sardonicus · 2020-03-21, 16:12

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.

LaurV · 2020-03-22, 09:46

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...

2020-03-21, 03:31	#1
Citrix Jun 2003 1582₁₀ 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

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2020-03-21, 15:18	#4
Citrix Jun 2003 2×7×113 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 11674₈ Posts	The closed form of S(n) looks to be (k^n-(-1)^n) / ((k+1)*k^n)

2020-03-22, 09:46	#7
LaurV Romulan Interpreter Jun 2011 Thailand 9611₁₀ 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...