Writing a recursive series explicitly?

Question

I am studying a recursive series which looks like this:

$n+\frac{n}{g} + \frac{(n+\frac{n}{g})}{g} + \frac{(n+\frac{n}{g} + \frac{(n+\frac{n}{g})}{g})}{g} + ...$

It is made of the terms in-between the + signs.

The series is recursive because every $n$ in the next term beyond $\frac{n}{g}$ is replaced by $n+\frac{n}{g}$. Since I want to sum these terms up to a certain index, I would need some way of writing them as a summation/sigma. How might I do this?

Answer 1

Apart from the first term that doesn't fit the pattern, you seem to be taking $\sum_{k=1}^M T_k(n)$ where $T_{j+1}(n) = T_j(n + n/g)$. Thus $$ \eqalign{T_1(n) &= \dfrac{n}{g}\cr T_2(n) &= \dfrac{n+n/g}{g} = n \dfrac{g+1}{g^2}\cr T_3(n) &= (n + n/g) \dfrac{g+1}{g^2} = n \dfrac{(g+1)^2}{g^3}\cr T_4(n) &= (n+n/g) \dfrac{(g+1)^2}{g^3} = n \dfrac{(g+1)^3}{g^4}\cr}$$ and in general $$ T_k(n) = n \dfrac{(g+1)^{k-1}}{g^k}$$ Then $$ \sum_{k=1}^M T_k(n) = \dfrac{n (g+1)^M}{g^M} - n $$

Answer 2

I will suppose, as Robert Israel pointed, that you have corrected your mistake and added another $n$ at the beginning of the series to make it coherent with your subsequent sentence.

Call $a_k(m)$ your $k$-th term where you have $m$ at the beginning

$a_0(n) = n$ and $a_{k+1}(n) = a_k(n(1+1/g))$.

Then $a_k(n) = a_{k-1}(n(1+1/g)) = a_{k-2}(n(1+1/g)^2) = \cdots = a_{0}(n(1+1/g)^k) = n(1+1/g)^k$.

So your series is $\sum_{k\geq 0} n(1+1/g)^k$.