I'm going through some past papers, and I'm asked to determine whether $$\sum_{k=1}^{\infty} e^{\left(\frac{1}{k^2}\right)}$$ converges or diverges. The question is fine, I'm just not sure about the level of rigour required. I cannot find any worked solutions that could give me an indication.
I have two variation of my solution:
1) As $k \rightarrow \infty$, it follows that $\frac{1}{k^2} \rightarrow 0$. This in turn means that $e^{\left(\frac{1}{k^2}\right)} \rightarrow 1$, which means as $k$ gets sufficiently large, $$\sum_{k=1}^{\infty} e^{\left(\frac{1}{k^2}\right)} = \sum_{k=1}^{N} e^{\left(\frac{1}{k^2}\right)} + \sum_{k=N+1}^{\infty} 1^{\left(\frac{1}{k^2}\right)}$$ Since the tail diverges, so does the series. $\square$
2) For all $k \in \mathbb{N}$, it follows that $$1^{\left(\frac{1}{k^2}\right)} < e^{\left(\frac{1}{k^2}\right)} \iff \sum_{k=1}^{\infty} 1^{\left(\frac{1}{k^2}\right)} < \sum_{k=1}^{\infty} e^{\left(\frac{1}{k^2}\right)}$$ Since $$\sum_{k=1}^{\infty} 1^{\left(\frac{1}{k^2}\right)} = \infty$$ it follows by the comparison test that $$\sum_{k=1}^{\infty} e^{\left(\frac{1}{k^2}\right)} = \infty$$ So the series diverges. $\square$
Is one solution better than the other? Is there an even better way to write a rigorous limit solution? Please give me some feedback.
It suffices to observe that
$$e^{\left(\frac{1}{k^2}\right)} \rightarrow 1$$
therefore the series doesn’t converge (it is a necessary condition that $a_k \to 0$).