Truncated sum of $e^x$

Question

I know that: $$\sum_{i=0}^\infty \frac{(x\ln{2})^i}{i!}=e^{x\ln{2}}=2^x $$ from the Taylor series for $e^x$. How can I find an approximation or asymptotic expression for the terminated sum: $$\sum_{i=0}^{x-1} \frac{(x\ln{2})^i}{i!}$$

I believe that this expression will be of the order $O(2^x)$, but am unsure how to proceed. Any help is appreciated.

EDIT: Sorry for the initial inconvenience. The upper limit for the sum is $x-1$ and not $t-1$.

By checking for large values of $x$ in Wolfram Alpha, it seems like: $$\sum_{i=0}^{x-1} \frac{(x\ln{2})^i}{i!} \sim 2^x$$

Answer 1

By the Stirling approximation, the last term is approximately

$$\frac{(x\log2\,e)^{x-1}}{\sqrt{2\pi(x-1)}(x-1)^{x-1}}\approx\frac{e}{\sqrt{2\pi x}}(\log2\,e)^{x-1}.$$

This is close to a geometric progression of common ratio $\log2\,e=1.884\cdots$ so that a reasonable approximation is

$$\frac c{\sqrt x}(\log2\,e)^{x-1}$$ where $c$ is a slowly varying function of $x$ which tends to

$$\frac e{\sqrt{2\pi}(\log 2\,e-1)}.$$

Answer 2

Use Taylor's series with remainder (form of Legendre):

$\begin{align*} f(x) &= \sum_{0 \le k \le n} \frac{f^{(k)}(0)}{k!} x^k + \frac{f^{(n +1)}(0)}{(n +1)!} \xi^{n +1} \end{align*}$

Here $0 < \xi < x$. Knowing the derivatives you can bound the remainder, and thus see when your estimate is accurate enough.

Answer 3

I claim that: $$\sum_{i=0}^{x-1} \frac{(x\ln{2})^i}{i!} \sim 2^x$$

To prove this, I am required to show that: $$\lim_{x \to \infty} \frac{1}{2^x}\sum_{i=0}^{x-1} \frac{(x\ln{2})^i}{i!}=1 \iff \lim_{x \to \infty} \frac{1}{2^x}\bigg(\sum_{i=0}^{\infty} \frac{(x\ln{2})^i}{i!}-\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!}\bigg)=1$$

We know by the Taylor series for $e^x$ that: $$\sum_{i=0}^{\infty} \frac{(x\ln{2})^i}{i!}=e^{x \ln{2}}=2^x$$

Now, we are required to show: $$\lim_{x \to \infty} \frac{1}{2^x}\bigg(2^x-\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!}\bigg)=1 \iff \lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!}=0$$

Note that for any $y \geqslant x$, we have: $$\frac{(x\ln{2})^{y+1}}{(y+1)!} \div \frac{(x\ln{2})^y}{y!}=\frac{x\ln{2}}{y+1}<\ln{2}$$

Thus, we can use a geometric progression to bound the sum:

$$\lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!} \leqslant \lim_{x \to \infty} \frac{1}{2^x}\sum_{j=0}^{\infty} \frac{(x\ln{2})^x}{x!} \cdot (\ln{2})^j=\lim_{x \to \infty} \frac{1}{2^x} \frac{(x\ln{2})^x}{x!} \cdot \frac{1}{1-\ln{2}}$$

We now use Stirling's Formula:

$$\lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!} \leqslant \lim_{x \to \infty} \frac{1}{2^x} \frac{(x\ln{2})^x \cdot e^x}{\sqrt{2\pi} \cdot x^{x+1/2}} \cdot \frac{1}{1-\ln{2}}$$

$$\lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!} \leqslant \lim_{x \to \infty} \frac{(e\ln{2})^x /2^x}{\sqrt{2\pi} \cdot x^{1/2}} \cdot \frac{1}{1-\ln{2}}=0$$

since $\frac{e \ln{2}}{2}<1$. Now, it is obvious that $$0 \leqslant \lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!}$$

since all the terms are positive. By Squeeze Theorem:

$$\lim_{x \to \infty} \frac{1}{2^x}\sum_{i=x}^{\infty} \frac{(x\ln{2})^i}{i!}=0$$

Hence, we have proved out conclusion:

$$\sum_{i=0}^{x-1} \frac{(x\ln{2})^i}{i!} \sim 2^x$$