Series Exercises

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Since

the divergence test says this series diverges.

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The divergence test is no help. The limit of the terms is zero:

This means there's hope for the series to converge and we need to try another test. This isn't a geometric series or an alternating series, it doesn't look like an integrable function, and it doesn't look very ratio-like (no exponents or factorials). That leaves the comparison test.

Since the denominator has a 3ⁿ we guess the series converges. We need to find a series with bigger terms that also converges. Thankfully, since making the denominator smaller makes the fraction bigger,

(we should point out that everything in sight is positive, so the required condition that the terms be non-negative is met). The series

converges (it's a geometric series with |r| < 1, or you could use the ratio test if you really wanted). This means the smaller series

also converges.

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Even though this really looks like a ratio test series, remember to try the divergence test first.

The formula for the nth term simplifies to

Now we can see that as n goes to infinity, the value of the nth term approaches . Since the terms don't converge to zero, the series diverges.

While we could use the ratio test, it's horrible and messy unless we simplify the terms first. And if we're going to simplify the terms anyway, we may as well use the divergence test.

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We try the divergence test first. Since the terms approach zero, all the divergence test tells us is that this series might converge.

It's not a geometric series or an alternating series. This looks like an integrable function, so integrate:

Since the function  is non-negative and decreasing on [1,∞), and its integral diverges, the series diverges too.

What about the two tests we didn't use? Could we have used either of them?

• We can't use the ratio test: so the ratio test tells us nothing.
• We can't use the comparison test: making the denominator smaller makes the fraction bigger, so . Since the harmonic series diverges, this comparison goes the wrong way. There aren't any other obvious series to compare to.

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The terms go to zero, and this looks like a ratio test series, so let's use the ratio test:

Since the limit of the ratio of the terms is less than 1, the ratio test says the series converges.

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The terms converge to zero. There are many right ways to go from here.

• The series is geometric with r = -0.9. Since |r| < 1, the series converges.
• This is an alternating series whose terms have strictly decreasing magnitudes and are approaching zero. Thus it converges by the AST.
• The ratio test gives us so the series converges.

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The divergence test says that since the terms converge to zero, the series might converge. Again, we have options.

• This is an alternating series whose terms are approaching zero and have strictly decreasing magnitudes. The AST says the series converges.
• We can show that the series of absolute values converges by either the integral test or the ratio test. Since absolute convergence implies convergence, the original series converges.

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The divergence test says that since the terms approach 0, there's hope for convergence. This isn't an alternating or geometric series. It's awfully messy, so the ratio test doesn't seem like a good idea. The comparison test doesn't sound like a great option either.

The integral test is the only one left, and conveniently the terms of the series have the form

which means they came from the ln function. It looks like the integral test is a good one to try. For positive x, the function

is non-negative and decreasing. We could do lots of work to show this or, easier, just look at a graph. Look at the indefinite integral of f:

Since the integral diverges, the series does too.

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The graph of arctan looks like this:

As n approaches ∞, arctan n approaches . This means

The divergence test says the series diverges.

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The denominator seems to have more n factors than the numerator, so maybe this series converges. This looks like a ratio test-able one.

We'll write out the factorial twice because squares of factorials are too complicated:

The ratio test says this series converges.