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# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = \ln(2n^2 + 1) - \ln(n^2 + 1)$

## The sequence converges to ln 2

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for this problem when we're looking at the limit is n goes to infinity of am We're definitely not allowed to just plug in and equals infinity because then we'd have infinity here and we'd be subtracting infinity. So infinity minus infinity is indeterminate form doesn't mean that this limit doesn't necessarily exist. It just means that you're not allowed to do that. So something else that you could do would be to use properties of logs to know that subtraction on the outside corresponds to division on the inside. So instead of subtracting these two logs, we can write as one log the log of two in square posts, one divided by in squared plus one and the natural log function is continuous. So we're allowed to pull the limit inside of the function. And then we do the trick where we look at the denominator and we look at the term that's going to infinity, the fastest, and then we divide the top in the bottom by whatever that happens to be. So in this case we divide the top in the bottom by n squared to get two plus one over n squared, divided by one plus one over N squared, as in goes to infinity. This and this are both going to go to zero. So we just get natural log of two over one, which is something as natural log of two. So we converge and we converge to the natural log too.

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