TITLE: How do you solve a recurrence with a summation function inside: $t(n) = 1 + \sum\limits_{j=0}^{n-1} t(j)$
QUESTION [5 upvotes]: Show that
$$t(n) = 1 + \sum_{ j=0}^{n-1} t(j)$$
is the same as
$$t(n) = 2^n$$
Initial condition $t(0) = 1$

REPLY [5 votes]: The rule to remember here is very simple: the generating function of the sum of the initial $n$ terms of a sequence is obtained by multiplying the generating function of that sequence by $1/(1-z).$
In your example we put $T(z) = \sum_{n\ge 0} t_n z^n$ and we have
$$t_n = 1 + [z^{n-1}] \frac{1}{1-z} T(z)$$
where $[z^n]$ is the coefficient extraction operator for formal power series.
Multiply by $z^n$ and sum over $n\ge 1$ to obtain
$$\sum_{n\ge 1} t_n z^n = 
\sum_{n\ge 1} z^n + \sum_{n\ge 1} z^n [z^{n-1}] \frac{1}{1-z} T(z).$$
Simplifying we obtain
$$T(z) - 1 = \frac{z}{1-z} + z \sum_{n\ge 1} z^{n-1} [z^{n-1}] \frac{1}{1-z} T(z).$$
This yields
$$ T(z)-1 = \frac{z}{1-z} + z \frac{1}{1-z} T(z).$$
Solve this for $T(z)$ to obtain
$$ T(z) = \frac{1}{1-2z}$$
so that by the geometric series $$t_n = 2^n.$$
Multiplication by $1/(1-z)$ is a very simple yet powerful tool when working with sums in recurrence relations.