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Integral test

# Worked example: Integral test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.6 (EK)
See how the integral test is put to use in determining whether a sequence converges or diverges.

## Want to join the conversation?

• I'm confused about part 2 of the integral test. In the last video, Sal showed that the series is an underestimate of the integral, and therefore a convergent (finite) integral would bound the series and make the series convergent too.
But it doesn't follow that just because the integral is divergent, then the series also diverges, because the integral is larger than the series.
• Actually, for part 2 you set it up so that the integral is smaller than the series. Therefore if the integral diverges, the corresponding series must diverge as well. It's all in how you draw your rectangles for the series (left endpoints or right endpoints) which allows you to visualize the series as being either less than or greater than a desired (and known) integral. Does that help?
• Usually the videos are very clear about everything, but this is the first time i was left a little unsure.
In the previous video ("Integral test intuition") we see Sal using the (1/x^2) function as an upper limit to the sum of (1/n^2) and showing how each block in the sum is less than its respective section in the integral. I am assuming that when n=1, the point on the graph is from ( n, (1/n^2) ) and drawn left to previous point. The boxes are clearly smaller and thus this "proof" is clear.

For this video however, without any shown reasoning - other than the function is now (1/x) and the sum is the harmonic series, 1/n, Sal draws from (n, 1/n) but to the right making the boxes bigger. Thus the sum is bigger and is pushed to divergence etc. If I hadn't accepted the rule that diverging(converging) function = diverging(converging) series respectively, then I would draw the boxes to the left (as in the previous film) making them smaller than the function and thus the whole test inconclusive.
I'm sure there is a small detail somewhere about where the integral or sum starts etc, but since i didn't notice it here, it will be a trap during an exam.
• I think that Sal isn't worried about that detail because the functions are so closely connected that it doesn't matter whether you overestimate or underestimate the function. Both the function and the estimation converge or diverge independent of the estimation. The only reason estimating over or under would matter would be to find the value that it approaches, but because they are both estimations that follow the function closely and do not present a number that's accurate, the integral itself is all that matters.
Sal does show some proof in the first video by comparing that sum to the integral plus the first value of the series. ∑ < ∑(1) + ∫ This allows comparison to an overestimate and allows a function that converges to be proven as convergent.
In the second video, Sal compares the sum directly to the integral ∑ > ∫ leaving the integral in a position where it is smaller than the sum and and where it is able to prove the sum divergent.
This could be done every time you use the integral test but the integral follows the series so closely that it would be a waist of time. Correct me if I'm wrong.
• In the sketches, how can you use left Riemann sum for 1/x and then right Reimann sum for 1/x^2? Isn't the Riemann sum open to interpretation, and not actually prove anything in this situation? My question is: why don't you be consistent with either the right or left hand Riemann sum?
• In the previous video titled "integral test intuition", Sal uses a right Riemann sum for 1/x^2 because it represents the sum of rectangles that lie entirely underneath the curve. The purpose of doing that was to argue that the sum of the first n terms minus the first term is less than the integral from 1 to n. And since the improper integral was convergent in that case, the series was bounded above by a positive number. And since we have an increasing series, that series must converge to some number.

In this video, Sal uses left Reimann sums because the rectangles lie above the curve. This again allows him to convincingly argue that the sum of the series(1/n) is divergent because the for any given n, the sum of the first n-1 terms is always greater than the integral between 1 and n. And since the integral of f is divergent, and f is always positive, the integral must be unbounded. That implies that the series is unbounded and must therefore diverge.

In short, Sal's decision to switch from a right to a left Riemann sum was strategic on his part because the left Reimann sums for 1/x always lie above the curve, making it easier for him to visually confirm that it does indeed diverge.
• What's the difference between this function and the one in the previous video (other than 1/x^2 and 1/x) where he drew the area of 1 from 0 to 1, but here he drew the area of 1 from 1 to 2?
• there is something strange with limits of int(1/x) and Int(1/x^2. looking at graphs of antiderivatives its obviuos that the first one diverges and the second one converges on (1, infinity). But if I observe graphs of d(lnx)/dx and d(-1/x)/dx and areas under those curves ( integrals we initially have ) , then it looks like both areas should converge. Farther from 1 is x, smaller f(x) and therefore area is increasing/decreasing with smaller and smaller portions, which eventually should lead to convergence. why does it seem to be true ? What am I getting wrong?
• If understand your question, your concern is that even though 1/x, and 1/x^2 have a similar looking shape when graphed, why then do their series not converge?

You cannot always tell weather the series of a function converges based on its graphed shape alone, which is why we need all the tests in this playlist. You must analyze the function, it's an unpleasant truth.
• Simply can anyone say me how series and integral are related?
• Do we have a simple and actual proof for this test?
• Sal sketches what the proof would look like in two parts, with part 1 (convergence of the integral implies convergence of the series) being in the video preceding this one and part 2 (divergence of the integral implies divergence of the series) being toward the end of this video. While the presentation isn't rigorous enough to qualify as a full proof, we should be able to understand from this explanation why this test has to be true.
• Why did khan start the series from 1 instead of 0 as he did in the previous video (Integral test intuition.) If the first term of the series starts from 0, then it seems like the series will be bounded by the integral of 1/x from 1 to infinity.
• I think it is because 1/x hits a vertical asymptote at x=0, which means it is going towards infinity in the vertical direction. and that makes testing this series considerably harder.