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

1. bahrom7893 Group Title

I loathe them:

2. bahrom7893 Group Title

this one and one more left.

3. Zarkon Group Title

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

4. Zarkon Group Title

|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 Group Title

TY TY TY TY TY Zarkon.

6. Zarkon Group Title

$\int_{n}^{n+1}\frac{1}{x}dx=\ln(n+1)-\ln(n)$ |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 Group Title

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

8. Zarkon Group Title

use you calculator or computer to get part (d)

9. bahrom7893 Group Title

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

10. Zarkon Group Title

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

11. bahrom7893 Group Title

ok sure one more and that's it then.