A logarithmic convergence tests - Analysis

[Wuberdall]

Homework Statement

Given a non-negative sequence \{a_{n}\}_{n=1}^{\infty}. Proove that the serie \Sigma_{n=1}^{\infty}a_{n} converge if and only if \Sigma_{n=1}^{\infty}\ln(1+a_{n}) converges.

Homework Equations

The Attempt at a Solution

My first attempt is the direct comparison test. Which tells that if \Sigma_{n=1}^{\infty}\ln(1+a_{n}) converges and if there exist a reel number γ so a_{n}\leq\gamma\ln(1+a_{n}) for all n, then the series \Sigma_{n=1}^{\infty}a_{n} also converges.

But i can't find such a γ.

 

[xiavatar]

Hint*: Note that $\sum_{n=1}^\infty ln(1+a_n)$ is a sum of logarithms. What can you do to change logarithms into multiplication?

*$log(a)+log(b)=log(ab)$.

 

[Wuberdall]

yeah, so \Sigma_{n=1}^{\infty}\ln(1+a_{n}) = \ln\Big[\Pi_{n=1}^{\infty}(1+a_{n})\Big]. But how am I supposed to conclude from this, that the series \Sigma_{n=1}^{\infty}a_{n} must then also converge ?

 

[xiavatar]

Well we know that $\sum_{n=1}^\infty ln(1+a_n)=C$, where C is a constant, since the sum converges. And we know $\sum_{n=1}^\infty ln(1+a_n)=ln[\prod_{n=1}^\infty(1+a_n)]=C$. So if we take the exponent of both sides we get $\prod_{n=1}^\infty(1+a_n)=e^C$. If you expand the product what do you notice? Now you have a bound for the sum $\sum_{n=1}^\infty a_n$

 

[pasmith]

yeah, so \Sigma_{n=1}^{\infty}\ln(1+a_{n}) = \ln\Big[\Pi_{n=1}^{\infty}(1+a_{n})\Big]. But how am I supposed to conclude from this, that the series \Sigma_{n=1}^{\infty}a_{n} must then also converge ?

Why is that if every a_n \geq 0, we have that \exp\left(\sum_{n=1}^N \ln(1 + a_n)\right) = \prod_{n=1}^N (1 + a_n) \geq 1 + \sum_{n=1}^N a_n \geq 1<br /> for every N \in \mathbb{N}? (Hint: consider what happens if you expand the product.)

If \sum_{n=1}^\infty \ln(1 + a_n) converges, what can you say about the convergence of \sum_{n=1}^\infty a_n?

Conversely, if \sum_{n=1}^\infty a_n diverges, what can you conclude about the convergence of \sum_{n=1}^\infty \ln(1 + a_n)?