<s>[INST] Evaluate exactly the following series that includes fibonacci How can i find the exact value of this?
$$\sum_{n=1}^∞ \frac{f_n}{100^n}$$
Assuming that $ f_1=1$ and $f_2=1$ and $f_n=f_{n-1}+f_{n-2}$
I can approximate it to 0.01010203050813.........
But what is the exact value? Should i use $\lim_{n\to ∞}$ ? And how to solve it?
Thank you.
 [/INST] Following the hint of @lulu above, we let
$$
F(x) = \sum_{n = 1}^\infty f_nx^n
$$
We are after $F(\frac1{100})$, and to do that, we need a good expression for $F(x)$. The Fibonacci recurrence relation gives
$$
F(x) = \sum_{n = 1}^\infty f_nx^n =  x + x^2 + \sum_{n = 3}^\infty f_nx^n\\
= x + x^2 + \sum_{n = 3}^\infty(f_{n-1} + f_{n-2})x^n\\
= x + x^2 + \left(\sum_{n = 3}^\infty f_{n-1}x^n\right) + \left(\sum_{n = 3}^\infty f_{n-2}x^n\right)
$$
To get a better grip on those two sums, note what happens when we divide them by $x$ and $x^2$ respectively:
$$
\frac1x\sum_{n = 3}^\infty f_{n-1}x^n = \sum_{n = 3}^\infty f_{n-1}x^{n-1} = \sum_{n=2}^\infty f_nx^n = F(x) - x\\
\frac1{x^2}\sum_{n = 3}^\infty f_{n-2}x^n = \sum_{n = 3}^\infty f_{n-2}x^{n-2} = \sum_{n = 1}^\infty f_{n}x^n = F(x)
$$
Each of these lines give us
$$
\sum_{n = 3}^\infty f_{n-1}x^n = x(F(x) - x)\\
\sum_{n = 3}^\infty f_{n-2}x^n = x^2F(x)
$$
Inserting this, we have
$$
F(x) = x + x^2 + x(F(x) - x) + x^2F(x)\\
F(x) = x + x^2 + xF(x) - x^2 + x^2F(x)\\
F(x) -xF(x) - x^2F(x) = x\\
F(x)(1-x-x^2) = x\\
F(x) = \frac{x}{1-x-x^2}
$$
</s>