Comparing Improper Integrals Involving Infinity

From my current understanding: $K>J$ and $L>K$ , therefore $L>K>J$. How can I compare the first integral $I$ ?

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Hint: $x < e^x$ so $I < \ldots$. –  Robert Israel Jan 16 '13 at 4:12

I might go about the original problem a different way. Of $J$, $K$, and $L$, there is only one where the antiderivative is easy to compute, and it's easy to see whether that one converges. That observation eliminates several of the 6 alternatives given. Of those remaining, only one is not nonsense (e.g., (f) is nonsense since it deduces convergence from being bigger than something that converges).