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Prove the series converges or diverges
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\[\sum_{n=1}^{\infty}(\sqrt[n]{2n^{2}}-1)\] Is there a nice direct comparison I can use for this? Ratio is inconclusive, and limit comparison test is inconclusive (based on the conditions of convergence we're allowed to use from our notes. Limit comparison for this will work using conditions in my calculus text, I just am not allowed them since they are not in what our class has for conditions of convergence for limit comparison). So given those are inconclusive, is there a nice direct comparison or am I forced to use the idea of cauchy sequences or something?
compare to 1/n
to show that \[\sqrt[n]{2n^{2}}-1\ge\frac{1}{n}\] it is equivalent to show that \[2n^2\ge\left(1+\frac{1}{n}\right)^n\]
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