Show that if $ a_n > 0 $ and $ \sum a_n $ is convergent, then $ \sum \ln(1 + a_n) $ is convergent.
were given that the Siri's is conversion and that the terms in the syriza positive. Let's show that this Siri's also converges. So here, let's use the Lim comparison test. So this requires that we look at the limit as n goes to infinity of the natural log one plus a M over this other term over here, which is just an now, let's go ahead and replace a N with Let's Do X equals one over and we know that the limit of a n equal zero. That's just using the divergence test since this Siri's convergence. So by diversion test. So since X equals one over a n, we must have a CZ and goes to infinity, that ex close to infinity because one over zero goes to infinity. So here I can replace this with Lim is X approaches infinity. And then how did we rewrite this? Well, so we have Ellen. This is one plus and that's one of Rex. But then on the denominator, that's just one over X. So here, as we take the limit, both numerator and denominator. Well, first of all, the numerator as X goes to infinity, this just goes to Ellen of one, but that zero. And in the denominator we have one over X. But that just goes to zero. So we have a limit of the form zero over zero. This is indeterminate form, so we use Lopez's house rule. So here we take a look. We're going to use Lopez House rules. I abbreviate that appear with the L. H. So we rewrite that limit. But then we take the derivative of the numerator and denominator. So in the numerator we have one over one plus one over X and then by the chain rule. We have this extra negative one over X squared and then on the denominator. He's the power rule there to differentiate one of Rex, and we could cancel the negative one over and square terms. And then we just take the limit of this. And that's just one over one plus zero, which is one. So since this Siri's converges, our Siri's will also convert violin a comparison test because our limit satisfies this inequality. It's a number that's bigger than zero, but less in infinity. No, c was one. In our case, this is what is allowing us to use the Lim comparison test. So also converges Bye, Living Computer Sim, and that's our final answer