Show that series $\Sigma_{n=1}^{\infty} k(e^{\frac{3}{k}}-1)$ converges or diverges.

Question

I cannot find a way to determine whether series: $$\Sigma_{n=1}^{\infty} k(e^{\frac{3}{k}} - 1) $$ diverges or converges. First checking for necessary condition: $$lim_{k \to \infty} k(e^{\frac{3}{k}}-1) = \left[ \frac{0}{0} \right] \stackrel{L. R.}{=} \frac{\frac{3}{k^2} e^{\frac{3}{k}}}{\frac{1}{k^2}} = 3 \Rightarrow series \ diverge$$

Answer 1

Yes, it diverges, but your argument is not correct. In fact,\begin{align}\lim_{k\to\infty}k\left(e^{\frac3k}-1\right)&=\lim_{k\to\infty}\frac{e^{\frac3k}-1}{\frac1k}\\&=\lim_{k\to\infty}\frac{-\frac3{k^2}e^{\frac3k}}{-\frac1{k^2}}\\&=\lim_{k\to\infty}3e^{\frac3k}\\&=3.\end{align}

Answer 2

$a_n=k(e^{3/k}-1)\gt$

$ k(1+3/k-1)=3;$

$\lim_{n \rightarrow \infty}a_n \not =0$, hence divergent.

Used: For x=k/3:

$e^x =1+x+x^2/2!+...>1+x.$