Can $e^x$ be expressed as a linear combination of $(1 + \frac x n)^n$? In other words, does there exist an infinite sequence $(a_k)_{k \in \mathbb N_0}$ such that $$e^x = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$$ for all $x \in \mathbb R$? Call the series on the right $s(x)$.

I can answer the question in the negative when the series is absolutely convergent. In the conditionally convergent case, I'm not so sure. My thoughts were to use the fact that:

$$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x}$$

and use the lower bound when $a_k$ is negative, and the upper bound when $a_k$ is positive. This gets stuck because it's not always the case that if some $b_k$ is a decaying sequence then $\sum_{k} \frac{b_k}{k}$ is convergent.

The strengthened inequality $$e^{x - \frac{x^2}{2k}} \leq (1+ \frac x k)^k \leq e^{x - \frac{x^2}{2k} + \frac{x^3}{3k^2}}$$ looks like it might make more progress...

[EDIT 2019/08/14 14:00 GMT]

**This is the solution in the absolutely convergent case, given by lemmas 1 and 2**:

Definition: Let $s(x) = a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k$.

Lemmas and proofs follow:

Lemma 1: If $s(x)$ converges absolutely forsome$x\geq 0$, then $s(x)$ converges absolutely forall$x \geq 0$.

### Proof

Pick an $x_0 \geq 0$ for which $s(x_0)$ converges absolutely.

By the condition stated in the lemma, the series $\sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x_0} k\right|^k$ must converge. We also observe that $|a_k| \leq |a_k| \left|1 + \frac {x_0} k\right|^k$ is true for all $k$. So by the Direct Comparison Test, the series $\sum_{0 \leq k < \infty} |a_k|$ must also converge. In other words, $s(0)$ is absolutely convergent.

Consider now any $x \geq 0$. The series $\sum_{0 \leq k < \infty} |a_k| e^{x}$ converges because it is equal to $e^{x} \sum_{0 \leq k < \infty} |a_k|$, which we proved to be convergent in the previous paragraph. We observe that $|a_k| \left|1 + \frac {x} k\right|^k \leq |a_k| e^{x}$ is true for all $k$. So by the Direct Comparison Test, the series $|a_0| + \sum_{1 \leq k < \infty} |a_k| \left|1 + \frac {x} k\right|^k$ must also converge. So by the definition of absolute convergence, we have that $a_0 + \sum_{1 \leq k < \infty} a_k \left(1 + \frac x k\right)^k=s(x)$ converges absolutely, where $x \geq 0$ was arbitrary.

$\blacksquare$

Lemma 2: If $s(x)$ converges absolutely when $x \geq 0$, then for large enough $x$ we have that $e^x > s(x)$.

### Proof

Let $z_n(x) = |a_0| + \sum_{1 \leq k < n} |a_k| \left(1 + \frac x k\right)^k$.

Pick some $\epsilon < \frac 1 2$.

Observe that there must be a large enough $n$ such that $z_\infty(0) - z_n(0) \leq \epsilon$.

Using the triangle inequality, we have that: $$\begin{aligned} |s(x)| &\leq z_\infty(x)\\ &\leq z_n(x) + (z_\infty(x) - z_n(x))\\ \end{aligned}$$

Since $z_n(x)$ is a polynomial, there is a large enough $X$ such that all $x \geq X$ it's true $z_n(x) < \epsilon \cdot e^x$. So we have that $$\begin{aligned} |s(x)| &<\epsilon\cdot e^x + (z_\infty(x) - z_n(x))\\ &\leq \epsilon\cdot e^x + (z_\infty(0) - z_n(0)) e^x\\ &\leq \epsilon\cdot e^x + \epsilon\cdot e^x\\ & = 2\epsilon \cdot e^x\\ &< e^x. \end{aligned}$$ The claim above that $z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x$ follows from $$\begin{aligned} &|a_k| \left(1 + \frac x k\right)^k \leq |a_k| e^x\\ \implies &\sum_{k \geq {n+1}}\left(1 + \frac x k\right)^k \leq \sum_{k \geq {n+1}}|a_k| e^x\\ \implies & z_\infty(x) - z_n(x) \leq (z_\infty(0) - z_n(0)) e^x \end{aligned}$$

We are done.

$\blacksquare$