Sequences and series

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Sequences and series

Mathematics
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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.

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I loathe them:
1 Attachment
this one and one more left.
we know know that \[\int_{1}^{n}\frac{1}{x}dx=\ln(n)\] ...

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

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