sequences and series - Does $sum_{n=1}^infty frac{1}{n!}{nbrace varphi(n)}$ converge, where ${nbrace m}$ are Stirling numbers of second kind?

When I was playing with the function

StirlingS2[n, m]

in Wolfram Alpha online calculator, that corresponds with the implementation of Stirling numbers of the second kind (see the definition in this MathWorld), I've thought about doing experiments with number-theoretic functions like the Euler's totient function $\varphi(n)$, the prime counting function $\pi(n)$ or the sum of divisor function $\sigma(n)$. With the purpose of ensuring the convergence of the series in my experiements I've multiplied by a factor $\frac{1}{n!}$ (and after I did my experiments I've seen that some series seems convergent, but other seems divergent).

I don't know if these kind of questions were in the literature, I am saying deduce the convergence of series with same form of next question (with $0StirlingS2[n, m]).

Question. Let $\varphi(n)$ the Euler's totient function and we denote with ${n\brace m}$ our Stirling numbers of second kind. Do you know how deduce the convergence of $$\sum_{n=1}^\infty \frac{1}{n!}{n\brace \varphi(n)}?$$ Thanks in advance.

Answer

We use an upper bound of the Stirling's number. A reference is formula (10) in here. This yields
$$
{n\brace \varphi(n)}\le \frac{ \varphi(n)^n}{\varphi(n)!}.
$$
Then we have

$$
\frac1{n!} {n\brace \varphi(n)} \le \frac{\varphi(n)^n}{n!\varphi(n)!}
$$
Stirling's formula gives
$$
\frac{\varphi(n)^n}{n!\varphi(n)!}\sim \frac1{\sqrt{2\pi n}}\left(\frac en\right)^n \varphi(n)^n \frac1{\sqrt{2\pi \varphi(n)}}\left(\frac e{\varphi(n)}\right)^{\varphi(n)}
$$
$$
\ll \frac{\exp(2n + (n-\varphi(n))\log \varphi(n) - n \log n)}{\sqrt{n\varphi(n)}}
$$

$$
\ll \frac{\exp(2n + n\log \frac{\varphi(n)}n - \varphi(n)\log\varphi(n))}{\sqrt{n\varphi(n)}}
$$
$$
\ll \frac{\exp(2n - \varphi(n)\log\varphi(n))}{\sqrt{n\varphi(n)}}
$$
Now, a series with this term converges because of this lower bound of $\varphi(n)$