sequences

[fading umbra]

can some one help me with how to derive the limit pls!

[pallid kiln]

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[fathom mountain]

Hm. Well, we could try solving the recurrence relation.

[fading umbra]

im not familiar with recurrence relation

i tried finding the nth term for the summation but i failed at it

im so confused if we even require require the nth term

[fathom mountain]

Let me show, then. You can search up the general algorithm if you want, it's pretty easy and doesn't require any advanced methods.
a(n) = (a(n - 1) + a(n - 2))/2, a(1) = 0, a(2) = 1
2a(n) - a(n - 1) - a(n - 2) = 0
The characteristic equation is:
2λ^2 - λ - 1 = 0
The roots are λ = -1/2 and λ = 1. So, the general solution is:
a(n) = A(-1/2)^n + B
Let's apply the initial conditions.
a(1) = -A/2 + B = 0
a(2) = A/4 + B = 1
The solution of that system is A = 4/3, B = 2/3. So, the solution is:
a(n) = (4/3)(-1/2)^n + 2/3
Or:
a(n) = (2/3)(1 - (-1/2)^(n - 1))
From here it's easy to find the limit.

[fading umbra]

the characteristic equation is something already predefined?

[fathom mountain]

Yes. The characteristic equation of Aa(n) + Ba(n - 1) + ... + Ua(n - k + 1) + Va(n - k) = 0 is Aλ^k + Bλ^(k - 1) + ... + Uλ + V = 0.

[fading umbra]

i think i get the grasp of it now

i will strengthen my knowledge further on this concept

thanks a lot for taking the time to explain!

[fathom mountain]

You're welcome!

Moreover, if you've dealt with linear differential equations before, the approach for linear recurrence relations will already be 95% familiar to you.

[fading umbra]

they still didnt introduce us linear differential equations

[fathom mountain]

Ah. Well, that's ok.

The approach still isn't terribly difficult.

[fading umbra]

yes its relatively easy