Is my claim that this number is irrational correct?

Question

Define a number $c$ in the following way: $$c=\ln \left(\prod^\infty_{k=1}\frac{e^{1/k}}{1+\frac{1}{k}}\right)$$ (I can assure you that this converges). Isn't this number transcendental since the input couldn't be a power of $e$? The denominator of the product is always an integer. It could be proven that: $$\prod^N_{k=1} 1+\frac{1}{k}=N+1$$ Since this is a telescoping product. We could make a limit for $c$: $$c=\lim_{N\rightarrow \infty}\ln \left(\frac{e^{H(N)}}{N+1}\right)$$ Where $H(n)$ is the $n$th harmonic number. $N$ approaches infinity in discrete steps because the product couldn't have a non-integer value on the top bound.

Answer 1

No, this kind of argument is not valid. It is true that the logarithms of the partial products

$$H_n - \log (n+1)$$

are transcendental. This follows from the (contrapositive of the) Lindemann-Weierstrass theorem, which implies more generally that if $\alpha \neq 1$ is a positive algebraic number then $\log \alpha$ is transcendental. But a limit of transcendental numbers isn't necessarily transcendental or even necessarily irrational; consider e.g. the sequence $\frac{\pi}{n}$ of transcendental numbers which approaches $0$.

In fact your constant $c$ is the Euler-Mascheroni constant $\gamma$ and it's an open problem whether it's irrational.

Answer 2

Well, $$c=\ln \left(\prod^\infty_{k=1}\frac{e^{1/k}}{1+\frac{1}{k}}\right)$$

is nothing but $\ln(e^{\gamma}) = \gamma$, where $\gamma$ is the Euler-Mascheroni constant.

If this constant is transcendental, is an open problem in Mathematics. Your argument here is invalid.