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bahrom7893
 3 years ago
Sequences and series
bahrom7893
 3 years ago
Sequences and series

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bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.3this one and one more left.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1we know know that \[\int_{1}^{n}\frac{1}{x}dx=\ln(n)\] ...

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1330845983664:dw the sum of the rectangles is the sum of 1/x which is larger than the area under the curve which is the integral thus \[\sum_{k=1}^{n}\frac{1}{k}>\ln(n)\] ie.. \[\sum_{k=1}^{n}\frac{1}{k}\ln(n)>0\]

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.3TY TY TY TY TY Zarkon.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1\[\int_{n}^{n+1}\frac{1}{x}dx=\ln(n+1)\ln(n)\] dw:1330846312070:dw the area of this rectangle is 1*(n+1)=n+1 and it is less than the area under the curve...thus... \[\frac{1}{n+1}<\ln(n+1)\ln(n)\] \[\frac{1}{n+1}\ln(n+1)<\ln(n)\] \[\sum_{k=1}^{n}\frac{1}{n}+\frac{1}{n+1}\ln(n+1)<\sum_{k=1}^{n}\frac{1}{n}\ln(n)\] \[a_{n+1}<a_{n}\]

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1for part c since \(a_n>0\) and \(a_{n+1}<a_n\) by the monotone bounded sequence theorem \(a_n\) converges.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1use you calculator or computer to get part (d)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.3TY SOO MUCH! One more problem left for today.. I hope, unless i get stuck on computational ones.

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1it is getting really late here so it needs to be quick if you want my help

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.3ok sure one more and that's it then.
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