with annual compounding:
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]$
with other compounding frequencies:
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
but that's probably what you referred to as sum of finite series... nevertheless, it's quite a short expression, I think, do you really need an integral here? Sorry for asking naively..
Derivation of the above:
First simplified, with annual payments.
Timeline:
$t = 0$ Lisa starts working
$t = 1$ Lisa gets her first annual salary, and makes the first payment into her savings account.
$t= T = 40$ Lisa retires. So she makes the last payment in $t=40$ (still one payment into the account from the last salary at the retirement date)
She works 40 years, makes 40 payments, and retires 39 years after the first payment.
The annual payment $P(t)$ is her savings rate $r=10%$ times her salary $S_t$ which is 40 K at the end of the first year, i.e. $S_1=40,000$ and then grows with $g=$5% per year, i.e.:
$P_t = S_1r(1+g)^{t-1}$
Payments into her savings account grow at an annual interest rate of $i=7$%, so the value at the retirement date ($T$) of a payment made in $t$ is:
$FV_{S_t}=S_tr(1+i)^{(T-t)}=S_1r(1+g)^{t-1}(1+i)^{T-t}$
The total value of the account is the sum over $t$ till $t=40$:
$\Sigma_tFV_{P_t}=$
$\Sigma_t[S_tr(1+i)^{(T-t)}]=$
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]=$
$\Sigma_t4,000(1.05)^{t-1}(1.07)^{40-t}$
with other compounding frequency:
number of compounding (sub)periods per year:
index for current subperiod (here months): $x$ $(1,2,...,12)$
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
Best Answer
Here's what we can do without finance functions. We know that in $2$ years, a trip will cost us $ \$ 9500$. We want to make a deposit every month so that, after $2$ years, our account will have $ \$ 9500$ in it. Let's call our undetermined deposit $X$. Note that $t$ is time in months. $i^{(12)}$ is the nominal interest, $i$ is the effective annual interest.
At $t=0$ we make no deposit.
At $t = 1$ we deposit $X$. Its accumulated value will be $X(1+ \frac{i^{(12)}}{12})^{23}$
At $t = 2$ we deposit $X$. Its accumulated value will be $X(1+ \frac{i^{(12)}}{12})^{22}$
...
At $t = 23$ we deposit $X$. Its accumulated value will be $X(1+ \frac{i^{(12)}}{12})^{1}$
At $t = 24$ we deposit $X$. Its accumulated value will be $X(1+ \frac{i^{(12)}}{12})^{0}$
We notice that this is a geometric series since we are summing the values of our deposits, so
$$\sum_{n=1}^{24} X \left(1+ \frac{i^{(12)}}{12} \right)^{24-n} = X \left( \frac{(1+ \frac{i^{(12)}}{12})^{24} - 1}{i} \right)$$
by some algebraic manipulations.
Then, we know that our calculated future value of the deposits has to equal the cost of the trip, so
$9500 = X \left( \frac{(1+ \frac{i^{(12)}}{12})^{24} - 1}{i} \right)$ from which it is easy to calculate $X$.
(whew).
Annuities lump this huge amount of work into a very concise notation with a simple formula that is extraordinarily flexible, so I recommend you read up on it. Any good interest theory book will go over it. I learned from Kellison's book and found it to be quite good.