Why does $\lim_{n \to \infty} \sum_{k=1}^n\frac{t^{k+1}}{(k+1)!}=e^t-t-1$?

Question

Why does $\lim_{n \to \infty} \sum_{k=1}^n\frac{t^{k+1}}{(k+1)!}=e^t-t-1$?

I know $\lim_{n \to \infty} \sum_{k=0}^n\frac{t^k}{k!}=e^t$, but my sum starts at $k=1$ and also has $\frac{t^{k+1}}{(k+1)!}$ instead of $\frac{t^k}{k!}$.

Answer 1

It's easier to shift indices, so that the summand looks the same between the sums you are comparing. Your sum is

$$\sum_{k=2}^{n+1} \frac{t^k}{k!}.$$

You can now "add and subtract" the first two terms of the sum for the exponential:

$$\sum_{k=2}^{n+1} \frac{t^k}{k!} = \sum_{k=0}^{n+1} \frac{t^k}{k!} - \sum_{k=0}^1 \frac{t^k}{k!} \\ = \sum_{k=0}^{n+1} \frac{t^k}{k!} - 1 - t.$$

Going up to $n+1$ is not really any different than going up to $n$ when you are sending $n \to \infty$, so you get your result.

Answer 2

$e^{t}=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+ \cdots$, then $$e^t-1-t=\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=\sum_{2}^{\infty}\frac{t^k}{k!}$$

Answer 3

Write out the first 5 or so terms of the series for the exponential and the series you have here. What you'll find is that the series you have here is exactly the series for the exponential but starting at the $t^2$ term. Thus if you subtract the first two terms from the exponential you get the series that you have here.

To transform one series into the other first note that $$e^t - t - 1 = \left(\sum_{k = 0}^\infty\frac{t^k}{k!}\right) - t - 1 = \sum_{k = 2}^\infty\frac{t^k}{k!}$$ Then shift variables by letting $r = k - 1$. You get $$\sum_{k = 2}^\infty\frac{t^k}{k!} = \sum_{r = 1}^\infty\frac{t^{r + 1}}{(r + 1)!}$$ Finally note that $r$ is just a dummy variable, so I can just rename it back to $k$. $$\sum_{r = 1}^\infty\frac{t^{r + 1}}{(r + 1)!} = \sum_{k = 1}^\infty\frac{t^{k + 1}}{(k + 1)!}$$

Answer 4

We have

$$\sum_{k=1}^n\frac{t^{k+1}}{(k+1)!}=\sum_{k=2}^{n+1}\frac{t^{k}}{k!}=\sum_{k=0}^{n+1}\frac{t^{k}}{k!}-1-t\xrightarrow{n\to\infty}e^t-1-t$$