How to find the series and its limit as n->INFINITY

Question

I have the first few number of a series and need to find the series and its limit:

I, for the life of me, cannot think of a way to represent this as a series.

Answer 1

\begin{align} 9t +8\sum_{k=2}^{n-1} \frac{t^k}{k!} -\frac{t^n}{n!} &=t +8\sum_{k=1}^{n-1} \frac{t^k}{k!} -\frac{t^n}{n!} t^n \\ &=t \color{red}{-8} +8\sum_{k=\color{red}{0}}^{n-1} \frac{t^k}{k!} -\frac{t^n}{n!} t^n \\ &\to t -8 +8 e^t\end{align} as $n \to \infty$.

Answer 2

The $n - th$ partial sum is given by

$s_n=9t+8(\frac{t^2}{2!}+....+\frac{t^{n-1}}{(n-1)!})-\frac{t^n}{n!}$, hence

$s_n \to 9t+8(e^t-1-t)-0=9t+8(e^t-1-t)=8e^t+t-8.$

Answer 3

An alternative approach - if we denote the nth function in the series by $f_n(t)$ then

$f_{n+1}(t) = \left[\int_0^t f_n(x) dx\right] + 9t - t^2/2$

If the series converges to a limit $f(t)$ then

$f(t) = \left[\int_0^t f(x) dx\right] + 9t - t^2/2$

so

$f'(t) = f(t) + 9 - t$

Solving this differential equation gives

$f(t) = ke^t + t - 8$

for some constant $k$. The additional constraint $f(0)=0$ gives $k=8$.