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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 6: Comparison tests

# Worked example: direct comparison test

Using the direct comparison test to determine that the infinite sum of 1/(2ⁿ+n) converges by comparing it to the infinite sum 1/2ⁿ.

## Want to join the conversation?

• how do you choose which series to compare the given series with? let's say, an = 1/((n^2)+2n). How do you choose which bn to compare with an?
• You need to find a series that is similar in behaviour to the one you are testing, yet simpler and that you know whether it converges or diverges.

In this case, your series `an = 𝚺 1/(n²+2n)` is pretty similar to `𝚺 1/n²` (which is a know convergent series), furthermore, the extra `2n` in the denominator of your series will make each term smaller than the corresponding term of our test series, so the convergence is assured.
• where did he get the if 1>1/2 it converges
• What if in this video we choose to use the harmonic series 1/n ? It is bigger than the firs series. The problem is, it diverges. So, why if I find a series that is bigger than converges, I say that my series converges; even though I know there is some other bigger series that diverges? I get the fact that you need the bigger series to converge to say that my series converges, and vice versa, I need the smaller series to diverge in order to being able to say that my series diverges.
My question is: I find two series that are bigger than my series. One diverges and one converges, or even, the smallest one diverges (even though it's bigger than my series) and the bigger one converges. (Maybe it's not possible and my question it's pointless). But what gives me the assurance that yes, I found ONE that converges, so MINE must converge to? It looks convenient, but not convincing.
• The test can only tell you these two things:
Suppose you have a series An.
If you find a CONVERGENT series Bn such that Bn>=An for all n, then An MUST ALSO CONVERGE.
Suppose you have a series An.
If you find a DIVERGENT series Bn such that Bn<=An for all n, then An MUST ALSO DIVERGE.
• There's all these tests to figure out whether or not a series converges/diverges. How do you get the "intuition" of knowing which test to use?
• practice topic:Direct comparison test

What does the below sentence actually mean?I tried a practice Question n the hint was-

"Because our given series is term-by-term greater than a convergent series, the direct comparison test does not apply.
So the direct comparison test is inconclusive"
• It just means that if we know that the "bigger" series diverges, that information is not enough to determine whether or not the "smaller" series will diverge using the Direct Comparison Test
• What rule/test did he use to determine (1/2)^n converges?
• In order to see the formula that he is referring to you need to rewrite (1/2)^n in the form ar^k. If you remember from an earlier video this then converges to a/(1-r) provided that -1<r<1.
With this in mind you can rewrite (1/2)^n in the form ar^k or 1*(1/2)^k the sum of which is a/(1-r) or 1/(1-1/2) which is 1. Or more simply, if you cover half the distance from where you are to where you want to be, eventually you will get there.
See the first paragraph in this video:Proof of infinite geometric series formula
(1 vote)
• what are the conditions i can't use the comparison test ?
• Suppose you have two series with terms a sub n and b sub n respectively. A few condition must be met in order to properly use the comparison test. First, the terms of these series must be positive. Second, a sub n must be less than or equal to b sub n. And finally, when the first two conditions are met, the following comparisons can be used to justify a conclusion regarding convergence and divergence:
(1) If the sum of b sub n converges, then the sum of a sub n converges.
(2) If the sum of a sub n diverges, then the sum of b sub n diverges.

Notice however that the following statements are not justified by the comparison test:
(3) If the sum of a sub n converges, then the sum of b sub n converges.
(4) If the sum of b sub n diverges, then the sum of a sub n also diverges.

• Which test is Sal using at to prove that the b sub n series converges?