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Mathematics 4 Online
OpenStudy (anonymous):

i seem to be getting this wrong and can't find out why. can someone help me and see where i went wrong?

OpenStudy (anonymous):

\[\frac{ 12^n }{ (n+1)(8^(2n+1) }\]

OpenStudy (anonymous):

use ration test to determine convergance \[a_(n+1)= 12^(n+1)/((n+2)(8^2n+2))\] \[\lim_{n \rightarrow \inf}\left| \frac{ 12^(n+1) }{ (n+2)(8^(2n+2) }*\frac{ (n+1)(8^(2x+1) }{ 12^n } \right|\] \[\lim_{n \rightarrow \inf}\frac{ 12(n+1) }{ 8(n+2) }\] =12/8

OpenStudy (anonymous):

so it should be divergent. but webwork says its the wrong answer. did i mess up somewhere and i can't see it?

OpenStudy (rsadhvika):

\(\large a_n = 8^{2n+1}\) becomes \(\large a_{n+1} = 8^{2n+3}\)

OpenStudy (rsadhvika):

once u correct that, it ends in 12/64 in think

OpenStudy (anonymous):

oh! thank you. i see it now. thank you very much!!

OpenStudy (rsadhvika):

you're wlcme :)

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