Let's see. The limit claim is pretty widely discussed on MSE, so I'm assuming you're willing to believe that the limit does approach $e$. Once you have that, you can look at the formula
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt}
$$
and do a little fiddling. Let $m = \dfrac{n}{r}$, so that $n = rm$. Then rewrite:
$$
A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt} = P \left(1 + \frac{1}{m}\right)^{mrt}= P \left(\left(1 + \frac{1}{m}\right)^m \right)^{rt}
$$
Now as $n \to \infty$, the thing inside the large parentheses approaches $e$, so you get
$$
A(t) = Pe^{rt}.
$$
As for the main limit, the usual approach is to say you want to find
$$
L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n
$$
but instead, you compute
\begin{align}
\ln L
&= \lim_{n \to \infty} \ln \left(1 + \frac{1}{n} \right)^n \\
&= \lim_{n \to \infty} n \ln \left(1 + \frac{1}{n} \right) \\
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n},
\end{align}
which you can evaluate with L'hopital's rule (take derivative of top and bottom, since both
go towards 0):
\begin{align}
\ln L
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n} \\
&= \lim_{n \to \infty} \frac{\frac{1}{1 + \frac{1}{n}}\left(\frac{-1}{n^2}\right)} {\frac{-1}{n^2}}\\
&= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}\\
&= 1.
\end{align}
Since the natural log of your limit is $1$, the limit itself must be $e$.$$$$$$
with annual compounding:
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]$
with other compounding frequencies:
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
but that's probably what you referred to as sum of finite series... nevertheless, it's quite a short expression, I think, do you really need an integral here? Sorry for asking naively..
Derivation of the above:
First simplified, with annual payments.
Timeline:
$t = 0$ Lisa starts working
$t = 1$ Lisa gets her first annual salary, and makes the first payment into her savings account.
$t= T = 40$ Lisa retires. So she makes the last payment in $t=40$ (still one payment into the account from the last salary at the retirement date)
She works 40 years, makes 40 payments, and retires 39 years after the first payment.
The annual payment $P(t)$ is her savings rate $r=10%$ times her salary $S_t$ which is 40 K at the end of the first year, i.e. $S_1=40,000$ and then grows with $g=$5% per year, i.e.:
$P_t = S_1r(1+g)^{t-1}$
Payments into her savings account grow at an annual interest rate of $i=7$%, so the value at the retirement date ($T$) of a payment made in $t$ is:
$FV_{S_t}=S_tr(1+i)^{(T-t)}=S_1r(1+g)^{t-1}(1+i)^{T-t}$
The total value of the account is the sum over $t$ till $t=40$:
$\Sigma_tFV_{P_t}=$
$\Sigma_t[S_tr(1+i)^{(T-t)}]=$
$\Sigma_t[S_1r(1+g)^{t-1}(1+i)^{T-t}]=$
$\Sigma_t4,000(1.05)^{t-1}(1.07)^{40-t}$
with other compounding frequency:
number of compounding (sub)periods per year:
index for current subperiod (here months): $x$ $(1,2,...,12)$
$\Sigma_t[\Sigma_x[\frac{S_1r(1+g)^{t-1}}{m}(1+\frac{i}{m})^{(m-x)}](1+\frac{i}{m})^{[m(T-t)]}]$
Best Answer
$11\%$ per year, compounded continuously is (approximately) equivalent to $11.6\%$ per year, compounded annually. If $i$ is the annual interest rate, the equivalent continuous rate is $\ln(1+i)$. The reason is simply that $$e^{\ln(1+i)}=1+i,$$ so that if you invest a dollar at a continuous rate of $\ln(1+i),$ at the end of a year, you have exactly what you would have had you invested the dollar at a rate of $i,$ compounded annually.
Of course, if you know the continuous rate $\delta$ and you want the equivalent annual rate $i,$ it's just $$i=e^\delta-1$$