EDIT : I found an alternative proof which seems correct. You do not have to read until bold letters since they are wrong. I recommend reading from "New Solution".
Show that there are no real solutions of $$1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k} = 0$$ where $$H_k = \sum_{i=1}^{k} \frac{1}{i}.$$
I managed to prove this, and I want to know if my proof is correct, and if there are any other (better) ways of proving it. (This is my first question and my English may be incorrect since English is not my first language.)
The overall proof is proving by contradiction :
let $$f(x) = 1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k}$$
and suppose a real number $a$ such that $f(a) = 0$. Since $f(x) \geq 1$ for $x \geq 0$, it is trivial that $a \lt 0$.
Just a little distribution :
$$\begin{align} f(x) &= 1 + \sum_{n = 1}^{\infty} \frac{x^n}{\prod_{k=1}^{n} H_k} = 1 + \sum_{n = 1}^{\infty} \left\{ \frac{x^{2n-1}}{\prod_{k=1}^{2n-1} H_k} + \frac{x^{2n}}{\prod_{k=1}^{2n} H_k}\right\} \\ \\ &= 1 + \sum_{n=1}^{\infty} \frac{x^{2n} + H_{2n} x^{2n-1}}{\prod_{k=1}^{2n} H_k} \end{align}$$
and
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \left\{ \frac{x^{2n-1}}{(2n-1)!} + \frac{x^{2n}}{(2n)!} \right\} = 1 + \sum_{n=1}^{\infty} \frac{x^{2n} + 2nx^{2n-1}}{(2n)!} .$$
Meanwhile, $$H_k = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{k} \lt k$$ and $$\prod_{k=1}^{2n} H_k \lt (2n)!.$$
Since $a \in \mathbb{R^-}$, $$a^{2n} + H_{2n} a^{2n-1} \gt a^{2n} + 2na^{2n-1}.$$
Therefore, $$0 = f(a) = 1 + \sum_{n=1}^{\infty} \frac{a^{2n} + H_{2n} a^{2n-1}}{\prod_{k=1}^{2n} H_k}$$ $$\gt 1 + \sum_{n=1}^{\infty} \frac{a^{2n} + 2na^{2n-1}}{(2n)!} = e^a \gt 0$$
and it is a contradiction.
Edit : According to the comments, I found out that the following identity can not be concluded right away since $a^{2n} + H_{2n} a^{2n-1}$ may be negative for some $a$.
So now we have to show the whole identity for all $n \in \mathbb{N}$ and $a \in \mathbb{R^-}$ such that $f(a) = 0$ :
$$ \frac{a^{2n} + H_{2n} a^{2n-1}}{\prod_{k=1}^{2n} H_k} \gt \frac{a^{2n} + 2na^{2n-1}}{(2n)!}$$
or prove a weaker statement :
$$ \frac{a^{2n} + H_{2n} a^{2n-1}}{\prod_{k=1}^{2n} H_k} \gt \frac{(a \ln t)^{2n} + 2n(a \ln t)^{2n-1}}{(2n)!}$$
where $t \gt 0$ since the contradiction still holds if $f(a) \gt t^a.$
Since $a^{2n-1} < 0$, the inequality we want to show is equivalent to the following inequalities :
$$\frac{a + H_{2n}}{\prod_{k=1}^{2n} H_k} < \frac{a (\ln t)^{2n} + 2n (\ln t)^{2n-1}}{(2n)!},$$
$$a < \frac{2n \cdot \prod_{k=1}^{2n} H_k \cdot (\ln t)^{2n-1} - (2n)! \cdot H_{2n}}{(2n)! - (\ln t)^{2n} \cdot \prod_{k=1}^{2n} H_k}$$
for all $a \in \mathbb{R^-}$ such that $f(a) = 0.$
My further attempt was trying to show that there exists $t \in \mathbb{R^+}$ such that $$\frac{2n \cdot \prod_{k=1}^{2n} H_k \cdot (\ln t)^{2n-1} - (2n)! \cdot H_{2n}}{(2n)! - (\ln t)^{2n} \cdot \prod_{k=1}^{2n} H_k} \geq 0$$ for all $n \in \mathbb{N}.$
Suppose $t < e$ for a weaker statement, then $\ln t < 1$ and $$(2n)! - (\ln t)^{2n} \cdot \prod_{k=1}^{2n} H_k > (2n)! - \prod_{k=1}^{2n} H_k > 0,$$
and we get to prove: $$2n \cdot \prod_{k=1}^{2n} H_k \cdot (\ln t)^{2n-1} - (2n)! \cdot H_{2n} \geq 0$$
or
$$\ln t \geq \left\{ \frac{(2n-1)!}{\prod_{k=1}^{2n-1} H_k} \right\}^{\frac{1}{2n-1}} =: F(n).$$
If there exists $M \in \mathbb{R}$ such that $F(n) < M$ for all $n \in \mathbb{N}$, such $t \in \mathbb{R^+}$ will exist and the inequality above will hold.
I am now struggling to prove this with the identities :
$$\ln n + \frac{1}{n} < H_n < \ln n + 1$$
for $n \in \mathbb{N}$, and
$$\ln k = \sum_{n=1}^{\infty} \frac{1}{n} \cdot \left( \frac{k - 1}{k} \right)^n$$
for $\frac{1}{2} < k \in \mathbb{R}.$
Bad News : The function $F(n)$ seems to diverge when $n \to \infty$, according to here.
New Solution
Note that $f$ is defined for all $x \in \mathbb{R}$ and if $f(a) = 0$ for $a \in \mathbb{R}$ then $a < 0$. Assume $a < 0$ exists.
For $n \in \mathbb{N}$,
$$\begin{align} \int_{a}^{0} \frac{x^{n}-a^{n}}{x-a} dx &= \int_{a}^{0} (x^{n-1} + ax^{x-2} + \cdots + a^{n-1}) dx \\ \\ &= \left[ \sum_{k=0}^{n-1} \frac{a^{k}}{n-k} x^{n-k} \right]_{a}^{0} = - a^{n} \sum_{k=1}^{n} \frac{1}{k} = - a^{n} H_{n} \end{align}$$
Using this result,
$$\begin{align} \int_{a}^{0} \frac{f(x)}{x-a} dx &= \int_{a}^{0} \frac{f(x) - f(a)}{x-a} dx \\ \\ &= \int_{a}^{0} \frac{1}{x-a} \left[ \sum_{n=1}^{\infty} \frac{x^{n}}{H_{1} H_{2} \cdots H_{n}} - \sum_{n=1}^{\infty} \frac{a^{n}}{H_{1} H_{2} \cdots H_{n}} \right] dx \\ \\ &= \int_{a}^{0} \frac{1}{x-a} \left[ \sum_{n=1}^{\infty} \frac{x^{n} - a^{n}}{H_{1} H_{2} \cdots H_{n}} \right] dx \\ \\ &= \sum_{n=1}^{\infty} \frac{1}{H_{1} H_{2} \cdots H_{n}} \int_{a}^{0} \frac{x^{n} - a^{n}}{x-a} dx \\ \\ &= - \sum_{n=1}^{\infty} \frac{a^{n} H_{n}}{H_{1} H_{2} \cdots H_{n}} \\ \\ &= -a - \sum_{n=2}^{\infty} \frac{a^{n} H_{n}}{H_{1} H_{2} \cdots H_{n}} \\ \\ &= -a - a \sum_{n=2}^{\infty} \frac{a^{n-1}}{H_{1} H_{2} \cdots H_{n-1}} \\ \\ &= -a - a \sum_{n=1}^{\infty} \frac{a^{n}}{H_{1} H_{2} \cdots H_{n}} \\ \\ &= - a \cdot \left[ 1 + \sum_{n=1}^{\infty} \frac{a^{n}}{H_{1} H_{2} \cdots H_{n}} \right] \\ \\ &= - a f(a) \\ \\ &= 0 \end{align}$$
Since $f(0) = 1$, $f(a) = 0$ and $\int_{a}^{0} \frac{f(x)}{x-a} dx = 0$, there exists real number $a_{1} \in (a, 0)$ such that $f(a_{1}) = 0$. (If not, then it is a contradiction since $f \ge 0$ on $(a, 0)$, not $f = 0$ everywhere since $f(0) = 1$, but $\int_{a}^{0} \frac{f(x)}{x-a} dx = 0$.)
Similarly, we can find $a_{n+1} \in (a_{n}, 0)$ such that $f(a_{n+1}) = 0$.
Now there are two ways to prove that this is impossible.
First, since $f$ is an entire function, its zero-set should not have accumulation points. However, we can find infinite zeros in the interval $(a, 0)$ and it is a contradiction.
Second, let the (partial) zero-set of $f$ be $S$ such that $S \subseteq (a, 0)$. Since $S$ is bounded above, there exists $s := \sup S$. Now we can find a sequence $\{ s_{n} \}_{n \in \mathbb{N}}$ in $S$, that converges to $s$. Since $f$ is continuous, $0 = \lim_{n \to \infty} f(s_{n}) = f(s)$.
Since $s$ is a zero of $f$, there exists $s' \in (s, 0)$ such that $f(s') = 0$. Since $s' \in S$, if $s \ne 0$ then it contradicts the fact that $s = \sup S$.
If $s = 0$, $f(s) = f(0) = 0$ because $s$ is a zero of $f$. However, it is trivial that $f(0) = 1$ and it is a contradiction.
Therefore, if $f(z) = 0$ then $z \in \mathbb{C} \, \backslash \, \mathbb{R}$.
