Question

Sequences and series

Answer 1

I loathe them:

Answer 2

this one and one more left.

Answer 3

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

...

Answer 4

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..

\[\sum_{k=1}^{n}\frac{1}{k}-\ln(n)>0\]

Answer 5

TY TY TY TY TY Zarkon.

Answer 6

\[\int_{n}^{n+1}\frac{1}{x}dx=\ln(n+1)-\ln(n)\]



the area of this rectangle is 1*(n+1)=n+1 and it is less than the area under the curve...thus...

\[\frac{1}{n+1}<\ln(n+1)-\ln(n)\]

\[\frac{1}{n+1}-\ln(n+1)<-\ln(n)\]
\[\sum_{k=1}^{n}\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)<\sum_{k=1}^{n}\frac{1}{n}-\ln(n)\]

\[a_{n+1}<a_{n}\]