inequalities using integration Q - alternating harmonic series (1 Viewer)

gamja · Mar 30, 2023

how to do 11a)?

tywebb · Mar 30, 2023

$\text{Suppose }x=-t$

$\text{Noting that }1-x^n=(1-x)(1+x+x^2+x^3+\cdots+x^{n-1})\text { then}$

$(1+t)(1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1})=1-(-t)^n$

$\therefore1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1}=\frac{1-(-t)^n}{1+t}=\frac{1}{1+t}+(-1)^{n+1}\frac{t^n}{1+t}$

cossine · Mar 30, 2023

gamja said:
View attachment 38092
how to do 11a)?

Have you tried the geometric series, or following along the lines of the geometric series proof?

tywebb · Mar 30, 2023

It's kind of the same.

$a+ar+ar^2+ar^3+\cdots+ar^{n-1}=\frac{a(1-r^n)}{1-r}$

$a=1,r=-t$

$1-t+t^2-t^3+\cdots+(-1)^{n-1}t^{n-1}=\frac{1-(-t)^n}{1--t}=\frac{1}{1+t}+(-1)^{n+1}\frac{t^n}{1+t}$

gamja · Mar 30, 2023

ok thanks for ur help! I did b) but how would you do c)? doesn't that statement imply

=0: can you solve this integral without using by parts?

edit: I think I would need to take the limit as t approached infinity of both sides because there is an unended ... - but I still need to learn how to integrate function

tywebb · Mar 30, 2023

inequalities using integration Q - alternating harmonic series (1 Viewer)

gamja

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