 In this video, we provide the solution to question number four from the practice final exam for math 1220 In which case we're given six infinite series here And we have to determine which of the following series are absolutely convergent Okay, so absolutely convergent means that if we take the absolute values of all the terms then it would be convergent in that situation So looking at the first one here the sum of one over the natural log of n well win The natural log when you are you're zero at one you're negative before that so starting at n equals two all the terms in this Series and the natural log here are already going to be positive. So one over natural log of n This is already positive here, but I should make mention that this series These fractions are going to be bigger than one over n So this series is going to be larger than the harmonic series, which is divergent So by the comparison test this one is a divergent series When we look at the second one here, it is an alternating series It is convergent by the alternating series test, but to be absolutely convergent We need to take the series of absolute values for which that just gives you the harmonic series Which like we mentioned a moment ago is divergent. So that's not the case there if you look at choice C We're just working by process of elimination. I can also notice that F is the harmonic series. So it's not that one When you look at choice C, this is a geometric series, but the common ratio is three halves Which is greater than one. So that'll be a divergent geometric series. You look at D here this sequence of factorials The factorial sequence is going to converge towards infinity not zero. So it's divergent by the test of divergence So by process of elimination We already have that the answer has got to be E because the other five cannot possibly be right But how could we convince ourselves? This is in fact absolutely convergent not just by process of elimination Well, if you take the absolute value of sine of n over n squared plus one Sine well, I mean because the denominator is already positive So the absolute value is really just playing a role in the numerator. Sine is negative sometimes positive other times, but regardless the absolute value of sine It's gonna be bounded below by zero but bounded above by one. So Our sequence is bounded above the absolute sequence I should say is bounded above by one over n squared plus one for which as n goes to infinity This is asymptotically the same thing as one over n squared for which now we're comparing ourselves to a P series which would be convergent by the p-test since p is equal to 2 So what we see here is that our series so this series as A p-series is convergent It's limit comparable to this series, which is convergent This series is larger than this series. I should say this series is larger than that series Which is the absolute series for this one in question here So by the comparison test this one would likewise be convergent and since the series of Absolute terms is convergent that means the series is absolutely convergent So we can find this out by process of elimination, but if you want a rigorous argument there Yes, this fact this series is in fact absolutely convergent