Ratio test with limsup vs lim Code Answer

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Could I prove that the ratio test still works using $limsup(frac{a_{n+1}}{a_n})$ instead of $lim(frac{a_{n+1}}{a_n})$?
I think for $limsup<1$ I could show that for $epsilon>0, N>1 limsup(frac{a_{n+1}}{a_n})<1-epsilon.$ From there I can solve that $lvert a_krvert<(1-epsilon)^{k-N}lvert a_Nrvert$ for $k>N$. Thus, by comparison test the left-hand side will converge absolutely. Is this enough to show I can use the ratio test with $limsup$?

Answer

You have the right ideas, but your proof could use just a little tweaking. Let $L = limsup |a_{n+1}/a_n|$. If $L < 1$, then for $epsilon := (1 – L)/2$, there exists $Nin Bbb N$ such that $|a_{n+1}/a_n| < L + epsilon$ for all $nge N$, i.e., $|a_{n+1}| < |a_n|(1+L)/2$ for all $n ge N$. Thus $|a_n| le [(1 + L)/2]^{n-N}|a_N|$ for all $nge N$. Since $L < 1$, $(1 + L)/2 < 1$, thus $sum_{n = 1}^infty [(1 + L)/2]^{n-N}$ converges. Hence, by the comparison test, $sum_{n = 1}^infty a_n$ converges absolutely.

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