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 Citrix 2020-03-21 03:31

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

:smile:

 Dr Sardonicus 2020-03-21 13:18

[QUOTE=Citrix;540343]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

:smile:[/QUOTE]Do you mean

S([b]k[/b]) = (1-S([b]k[/b] -1))/k

What is the formula for the [b]k[/b]th term?
Show that for large values of [b]k[/b] the [b]k[/b]th term converges on 1/(k+1) for k>1

?

 axn 2020-03-21 15:12

[QUOTE=Dr Sardonicus;540366]Do you mean
[/QUOTE]

[QUOTE=Citrix;540343]For a given integer k[/QUOTE]

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=Citrix;540373]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[/QUOTE]

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... :blush:

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