Does $\sum_{n=1}^\infty \left(e^{\frac{1}{n}} - 1\right)$ converge? Evaluate $\sum_{n=1}^\infty \left(e^{\frac{1}{n}} - 1\right)$
$ \lim e^{\frac{1}{n}} = 1$
$\sum_{n=1}^\infty \left(e^{\frac{1}{n}} - 1\right) > \sum_{n=1}^\infty(-1) = -\infty$
I'm not sure how to solve it. I tried two no convergence ways. what do u guys think about it?
 A: Since $e^x\geq 1+x$ we have $e^{1/n}-1\geq 1/n$ and hence the sum diverges by the comparison test.
A: Your idea of using a comparison is a good one, however, the comparison $$e^{\frac{1}n}-1>-1$$
is not particularly helpful, since the sum over $-1$ diverges to $-\infty$, so the comparison doesn't actually tell us anything. A more useful comparison would be
$$e^{\frac{1}n}-1>\frac{1}n$$
since $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ and if $x$ is positive, this is just $1+x+\text{positive terms}$ so $e^x>1+x$.
In general, notice that $f(x)=e^x-1$ is a function which has $f(0)=0$ and $f'(0)\neq 0$. One can use the bounds provided by the derivative to show that $\sum_{i=0}^{\infty}f(a_i)$ converges absolutely exactly when $\sum_{i=0}^{\infty}a_i$ converges absolutely.
A: In THIS ANSWER, I showed using standard, non-calculus based tools that the exponential function satisfies the inequality
$$e^{x}\ge 1+x$$
Therefore, we have with $x=1/n$
$$e^{1/n}-1\ge \frac1n$$
Since the harmonic series, $\sum_{n=1}^\infty \frac1n$, diverges, then by the Comparison Test, the series of interest diverges.
