Given $a_n = {(1+\frac{1}{n})}^{1/n}-1$, does $\sum_{n=1}^∞ a_n$ converge? I am a calculus student, And I'm trying to find out If the following series converges.
Given
$a_n = {(1+\frac{1}{n})}^{1/n}-1$
Does the following series converges?
$\sum_{n=1}^∞ a_n$
My thoughts:
It looks like $a_n$ converges to $0$ very quickly as $n$ goes to infinity, So my intuition says yes.
I've been trying to show convergence for some time now, and didn't had much progress. No candidates for comparison test, Other theorems did not work either.
Any hints/Tips for beginners?
 A: According to the Cauchy convergence criterion of the series:
For $ \forall \epsilon>0,\exists N>0, \forall n>N$, for $\forall p \in \mathbb N_{+}$，
\begin{align}\left|a_{n+1}+a_{n+2}+\cdots+a_{n+p}\right| &=\left|\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}-1+\left(1+\frac{1}{n+2}\right)^{\frac{1}{n+2}}-1\right. \\  & \qquad\left. +\cdots+\left(1+\frac{1}{n+p}\right)^{\frac{1}{n+p}}-1 \right|\\ &<\varepsilon \tag{1}
\end{align}
If this condition is satisfied, then $\sum_{n=1}^{\infty} a_{n}$ converges.

*

*$\lim_{n \to \infty}a_n=0$
$$\begin{aligned} & \because \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1 \\ \because & \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} e^{\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}} \\ & \lim _{n \rightarrow \infty} \ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \ln \left(1+\frac{1}{n}\right) \\ &\qquad\qquad\qquad\qquad\; =\lim _{t \rightarrow 0} t \ln (1+t)=0 \\ \therefore & \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}=1, \lim _{n \rightarrow \infty} a_{n}=1-1=0 \end{aligned}$$


*$a_n=\left(1+\frac{1}{n}\right)^{\frac{1}{n}}$ is monotonically decreasing
Using the inequality $(2)$:
$$\quad \frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)=\ln (n+1)-\ln n<\frac{1}{n} \quad(n=1,2, \cdots) \tag{2}$$
$$\because \ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\frac{1}{n} \ln \left(1+\frac{1}{n}\right),$$
$$\therefore \quad \frac{1}{n(n+1)}<\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}<\frac{1}{n^{2}}$$
$$\because \quad \ln \left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}=\frac{1}{n+1} \ln \left(1+\frac{1}{n+1}\right)=\frac{1}{n+1}[\ln (n+2)-\ln (n+1)]$$
$$\frac{1}{n+2}<\ln (n+2)-\ln (n+1)<\frac{1}{n+1}$$
$$ \therefore \quad \ln \left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}<\frac{1}{(n+1)^2}<\frac{1}{n(n+1)}<\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}$$
So, $$\left(1+\frac{1}{n}\right)^{\frac{1}{n}}>\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}.\tag{3}$$


*So the right end of equation $(1)$ can be deflated into：
\begin{align}\left|a_{n+1}+a_{n+2}+\cdots+a_{n+p}\right| &=\left|\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}-1+\left(1+\frac{1}{n+2}\right)^{\frac{1}{n+2}}-1 \right. \\ & \qquad \left. +\cdots+\left(1+\frac{1}{n+p}\right)^{\frac{1}{n+p}}-1 \right|\\ &< p \left|\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1 \right|\\&<\varepsilon \tag{4}
\end{align}
($n$ is sufficiently large to satisfy $\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1<\frac{\varepsilon}{p}$.)

Since the limit $a_n$ is convergent, equation $(4)$ holds, so equation $(1)$ holds, so the series satisfies the Cauchy convergence criterion, so the series converges.
