If $(S_n)$ converges, then $(\ln \sigma_n )$ converges as well

Question

Let $(a_n)$ be sequence of positive integers and define

$$ S_n = \dfrac{1}{a_1} + ... + \dfrac{1}{a_n} $$

and

$$ \sigma_n = \left( 1 + \dfrac{1}{a_1} \right) ... \left( 1 + \dfrac{1}{a_n} \right) $$

To prove: $(S_n) $ converges $\implies$ $( \ln \sigma_n ) $ converges.

try:

Note that $\ln \sigma_n = \sum_{i=1}^n \ln \left( 1 + \dfrac{1}{a_i} \right) $

Suppose $S_n \to S$, then for any $\epsilon > 0$ take $N> 0$ so that $n>N$ implies

$$ |S_n - S| < \epsilon $$

We know that for $x>1$ $ln(x+1) < x $. Thus,

$$ |\ln \sigma_n -S| < |S_n - S| < \epsilon $$

So that $\ln \sigma_n \to S$ converges as well.

Is this argument correct?

Answer 1

Start by reading my comment below.

I think there are a couple typos in your solution but you have the right initial idea. You already said:

$\ln \sigma = \sum_{i=1}^\infty \ln \left( 1 + \dfrac{1}{a_i} \right) $

Now we can prove that $ln(x+1) \leq x$ in the domain [0, 1] (this should be done formally) by saying that when $x = 0$ both are the same and that the derivative of $f(x) = x$ which is $\frac{df}{dx} = 1$ is always greater than the derivative of $g(x) = ln(x+1)$ which is $\frac{dg}{dx} = \frac{1}{x+1}$ on the domain $x\in[0, 1]$. Thus we can use the comparison test for series:

$ln(x+1) \leq x \implies \sum_{i=1}^\infty \ln \left( 1 + \dfrac{1}{a_i} \right) \leq \sum_{i=1}^\infty \dfrac{1}{a_i}$.

Thus if $\sum_{i=1}^\infty \dfrac{1}{a_i}$ converges then so does $\sum_{i=1}^\infty \ln \left( 1 + \dfrac{1}{a_i} \right)$.

Answer 2

Your proof is not correct

$$\ln(\sigma_n)<S_n$$ $$\implies \ln(\sigma_n)-S<S_n-S<0$$ but we cannot take the absolute value.

Let $$b_n=\frac{1}{a_n}$$

$$(S_n) \; converges $$ $$\; \implies \sum b_n \; converges $$ $$\implies \;\lim_{n\to+\infty}b_n=0$$

$$\implies \ln(1+b_n)\sim b_n$$

$$\implies \sum ln(1+b_n) \; converges$$

$$\implies \;( \sigma_n) \; converges$$