While Studying the convergence of $\sum_{n=1}^{\infty} e^{1/n}-1$, i was pretty sure it converged because $\lim_{n \to \infty} e^{1/n}$ is one (!?). After taking the comparison test with the harmonic series, i note the $\lim_{n \to \infty} \frac{e^{1/n}-1}{\frac{1}{n}}$ is one, so the previous series should diverge. What is the error i made along the way?
Thank you.
Note that $e^x>1+x$, hence $e^{1/n}>1+\frac1n$, then we have
$$\sum_{n=1}^{\infty} (e^{1/n}-1)>\sum_{n=1}^{\infty} (1+\frac1n-1)=\sum_{n=1}^{\infty} \frac1n\longrightarrow\infty$$
So the series is divergent by comparison test.
you missunderstood the property that If an series convergent,the sequence must be convergent to 0.This property shows the necessary condition of an convergent series,but not the sufficent one.
So If you want to know whether it converges,you should find any other ways,as comparing with the harmonic series.
The number $e$ is defined as the limit of the increasing sequence $(1+{1\over n})^n.$ Hence $e>(1+{1\over n})^n,$ which implies $e^{1/n}>1+{1\over n}.$ Thus $e^{1/n}-1>{1\over n},$ hence the series is divergent due to the comparison test.
