# A logarithmic convergence tests - Analysis

1. May 8, 2014

### Wuberdall

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

Last edited: May 8, 2014
2. May 8, 2014

### 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?

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

3. May 8, 2014

### 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 ?

4. May 8, 2014

### 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 $\sum_{n=1}^\infty a_n$

5. May 8, 2014

### pasmith

Why is that if every $a_n \geq 0$, we have that $$\exp\left(\sum_{n=1}^N \ln(1 + a_n)\right) = \prod_{n=1}^N (1 + a_n) \geq 1 + \sum_{n=1}^N a_n \geq 1$$ 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 $\sum_{n=1}^\infty \ln(1 + a_n)$?

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