I did this: Limit comparison test with $\frac{1}{n}$

$$\lim_{n\to\infty} \frac{\frac{1}{n^{1+\frac{1}{n}}}}{ \frac{1}{n}}$$ simplify $$\lim_{n\to\infty} n^{-2-\frac{1}{n}}$$ I then concluded that because $\frac{1}{n}$ in the exponent approaches 0 as $n$ approaches $\infty$ then the whole limit converges to a finite positive number. So I said that because $1/n$ diverges, by LCT the original series also diverges

Can someone confirm?

  • 2
    $\begingroup$ Should be $n^{-1/n}$, not $n^{-2-1/n}$. $\endgroup$
    – vadim123
    Commented Nov 28, 2017 at 23:04
  • $\begingroup$ I see what I did wrong.. $\endgroup$
    – Smit Shah
    Commented Nov 28, 2017 at 23:06


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