Here's the question you clicked on:
bahrom7893
Sequences and series
this one and one more left.
we know know that \[\int_{1}^{n}\frac{1}{x}dx=\ln(n)\] ...
|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... \[\frac{1}{n+1}<\ln(n+1)-\ln(n)\] \[\frac{1}{n+1}-\ln(n+1)<-\ln(n)\] \[\sum_{k=1}^{n}\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)<\sum_{k=1}^{n}\frac{1}{n}-\ln(n)\] \[a_{n+1}<a_{n}\]
for part c since \(a_n>0\) and \(a_{n+1}<a_n\) by the monotone bounded sequence theorem \(a_n\) converges.
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.