Integral Test

Integral Test - Integral Test The last topic that we...

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Integral Test The last topic that we discussed in the previous section was the harmonic series. In that discussion we stated that the harmonic series was a divergent series. It is now time to prove that statement. This proof will also get us started on the way to our next test for convergence that we’ll be looking at. So, we will be trying to prove that the harmonic series, diverges. We’ll start this off by looking at an apparently unrelated problem. Let’s start off by asking what the area under is on the interval . From the section on Improper Integrals we know that this is, and so we called this integral divergent (yes, that’s the same term we’re using here with series….). So, just how does that help us to prove that the harmonic series diverges? Well, recall that we can always estimate the area by breaking up the interval into segments and then sketching in rectangles and using the sum of the area all of the rectangles as an estimate of the actual area. Let’s do that for this problem as well and see what we get. We will break up the interval into subintervals of width 1 and we’ll take the function value at the left endpoint as the height of the rectangle. The image below shows the first few rectangles for this area.
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So, the area under the curve is approximately,
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Now note a couple of things about this approximation. First, each of the rectangles overestimates the actual area and secondly the formula for the area is exactly the harmonic series! Putting these two facts together gives the following, Notice that this tells us that we must have, Since we can’t really be larger than infinity the harmonic series must also be infinite in value. In other words, the harmonic series is in fact divergent. So, we’ve managed to relate a series to an improper integral that we could compute and it turns out that the improper integral and the series have exactly the same convergence. Let’s see if this will also be true for a series that converges. When discussing the Divergence Test we made the claim that converges. Let’s see if we can do something similar to the above process to prove this. We will try to relate this to the area under is on the interval . Again, from the Improper Integral section we know that,
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and so this integral converges. We will once again try to estimate the area under this curve. We will do this in an almost identical manner as the previous part with the exception that we will instead of using the left end points for the height of our rectangles we will use the right end points. Here is a sketch of this case, In this case the area estimation is,
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This time, unlike the first case, the area will be an underestimation of the actual area and the estimation is not quite the series that we are working with. Notice however that the only difference is that we’re missing the first term. This means we can do the following, Or, putting all this together we see that,
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This note was uploaded on 11/10/2011 for the course MATH 136 taught by Professor Prellis during the Fall '08 term at Rutgers.

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Integral Test - Integral Test The last topic that we...

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