Ratio test with limsup vs lim Code Answer

Hello Developer, Hope you guys are doing great. Today at Tutorial Guruji Official website, we are sharing the answer of Ratio test with limsup vs lim without wasting too much if your time.

The question is published on by Tutorial Guruji team.

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.

We are here to answer your question about Ratio test with limsup vs lim - If you find the proper solution, please don't forgot to share this with your team members.