an>0, sum[n=1,infty][an] conv.: what about sum of ln(1+an)?

[evaeswari87]

if an>0 and the summation of n=1 to infinity an is a convergent series,pls give your comments about the convergence of summation n=1 to infinity ln (1 + an). justify your comments.

i have no idea on how to this question.pls give me a hint on how to solve this question.thank u

 

[stapel]

i have no idea....

If you honestly "have no idea" what is be talked about here, then I'm afraid you would need weeks or months of tutoring in order to get caught up on the material -- so I hope you're exaggerating. :shock:

Since the terms a[sub:u4wk48s5]n[/sub:u4wk48s5] are all positive, then the terms (1 + a[sub:u4wk48s5]n[/sub:u4wk48s5]) are all greater than 1, so what can you say about the terms ln(1 + a[sub:u4wk48s5]n[/sub:u4wk48s5]), from what you learned back in algebra?

What does it mean, according to the definition, for the sum of the series {a[sub:u4wk48s5]n[/sub:u4wk48s5]} to be "convergent"?

What are your thoughts? Please be complete. Thank you!

Eliz.

 

[pka]

Napier’s Inequality states: \(\displaystyle 0 < a < b \Rightarrow \quad \frac{1}{b} < \frac{{\ln (b) - \ln (a)}}{{b - a}} < \frac{1}{a}\).

Using that we can see that \(\displaystyle 1 < x \Rightarrow \quad \ln (x) < x - 1\).

Apply that to your problem using direct comparison.

 

[PAULK]

if an>0 and the summation of n=1 to infinity an is a convergent series,pls give your comments about the convergence of summation n=1 to infinity ln (1 + an). justify your comments.

i have no idea on how to this question.pls give me a hint on how to solve this question.thank u

I believe there is a Theorem on Infinite Products: It says that if the

SUM ( a[n] ) converges, then the

PRODUCT (1 + a[n]) also converges.

Now ln(1 + a[1]) + ln (1 + a[2]) + .... =

ln{ (1 + a[1])(1 + a[2])(....) } =

ln PROD (1 + a[n])

I think that will show convergence.

 

[evaeswari87]

thanks alot for all the help and i manage to solve the question.thank u once again