## bahrom7893 Group Title Sequences and series 2 years ago 2 years ago

1. bahrom7893

I loathe them:

2. bahrom7893

this one and one more left.

3. Zarkon

we know know that $\int_{1}^{n}\frac{1}{x}dx=\ln(n)$ ...

4. 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.. $\sum_{k=1}^{n}\frac{1}{k}-\ln(n)>0$

5. bahrom7893

TY TY TY TY TY Zarkon.

6. 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... $\frac{1}{n+1}<\ln(n+1)-\ln(n)$ $\frac{1}{n+1}-\ln(n+1)<-\ln(n)$ $\sum_{k=1}^{n}\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)<\sum_{k=1}^{n}\frac{1}{n}-\ln(n)$ $a_{n+1}<a_{n}$

7. Zarkon

for part c since $$a_n>0$$ and $$a_{n+1}<a_n$$ by the monotone bounded sequence theorem $$a_n$$ converges.

8. Zarkon

use you calculator or computer to get part (d)

9. bahrom7893

TY SOO MUCH! One more problem left for today.. I hope, unless i get stuck on computational ones.

10. Zarkon

it is getting really late here so it needs to be quick if you want my help

11. bahrom7893

ok sure one more and that's it then.