bahrom7893
  • bahrom7893
Sequences and series
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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bahrom7893
  • bahrom7893
I loathe them:
1 Attachment
bahrom7893
  • bahrom7893
this one and one more left.
Zarkon
  • Zarkon
we know know that \[\int_{1}^{n}\frac{1}{x}dx=\ln(n)\] ...

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Zarkon
  • Zarkon
|dw: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
  • bahrom7893
TY TY TY TY TY Zarkon.
Zarkon
  • Zarkon
\[\int_{n}^{n+1}\frac{1}{x}dx=\ln(n+1)-\ln(n)\] |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}
Zarkon
  • Zarkon
for part c since \(a_n>0\) and \(a_{n+1}
Zarkon
  • Zarkon
use you calculator or computer to get part (d)
bahrom7893
  • bahrom7893
TY SOO MUCH! One more problem left for today.. I hope, unless i get stuck on computational ones.
Zarkon
  • Zarkon
it is getting really late here so it needs to be quick if you want my help
bahrom7893
  • bahrom7893
ok sure one more and that's it then.

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