## More Examples on Series Problem 1: Test for convergence .

Answer: Since we have a power n in the series, we will use the Root-Test. Set .

We have .

Since ,

and ,

we get .

But, .

Hence, .

Therefore, by the Root-Test, the series is divergent.

Problem 2: Test for convergence .

Answer: The sum of two series converges, if both of the sums converge. Hence the series  will diverge, if we can show that  diverges, while the series  converges. Since = ,

and   and the series  diverges by the p-test, we conclude that  diverges.

On the other hand,  converges by the ratio test:  =  = 0 < 1.

This establishes that  diverges.

Problem 3: Test for convergence .

Answer: We will use the Ratio-Test (try to use the Root-Test to see how difficult it is). Set .

We have .

Algebraic manipulations give ,

since .

Hence, we have ,

which implies .

Since , we conclude, from the Ratio-Test, that the series is convergent.

Problem 4: Determine whether the series is convergent or divergent. .

It is easy to check that f(x) is decreasing on . Hence, for any , we have for any , ,

which implies ,

that is, .

Using this inequality, we get ,

since .

Since ,

we deduce that the partial sums associated to the series are not bounded. Therefore, the series is divergent.

Remark: Note that the proof given above is the proof of the Integral-Test. In other words, we may have just used to Integral-Test to get the conclusion. Also, the series given here is part of a type of series called Bertrand series defined as . [Trigonometry] [Calculus]
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