# Calculus/Sequences and Series/Exercises

The following exercises test your understanding of infinite sequences and series. You may want to review that material before trying these problems.

Each question is followed by a "Hint" (usually a quick indication of the most efficient way to work the problem), the "Answer only" (what it sounds like), and finally a "Full solution" (showing all the steps required to get to the right answer). These should show up as "collapsed" or "hidden" sections (click on the title to display the contents), but some older web browsers might not be able to display them correctly (i.e., showing the content when it should be hidden). If this is true for your browser (or if you're looking at a printed version), you should take care not to "see too much" before you start thinking of how to work each problem.

## Sequences

editConsider the infinite sequence

- Is the sequence monotonically increasing or decreasing?

Algebraically, since

we see that for . That is, starting from the second term, the sequence is strictly decreasing. It is easy to check how the first two terms compare by just plugging in and :

The first two terms are equal and thereafter the terms are strictly decreasing. Therefore, the sequence is monotonically decreasing.

Using calculus,

- Is the sequence bounded from below, from above, both, or neither?

- Does the sequence converge or diverge?

To show this directly, consider the limit

The two limits are equal by L'Hôpital's Rule, since the numerator and denominator of the expression in the first limit both grow to infinity.

Since the limit exists, it is the number to which the sequence converges.## Partial sums

editAssume that the *n*^{th} partial sum of a series is given by

- Does the series converge? If so, to what value?

- What is the formula for the
*n*^{th}term of the series?

Note that the series turns out to be geometric, since

## Sums of infinite series

editFind the value to which each of the following series converges.

and so is geometric with first term and common ratio . So

*n*

and so is geometric with common ratio and first term . Thus

## Convergence and divergence of infinite series

editDetermine whether each of the following series converges or diverges. (*Note:* Each "Hint" gives the convergence/divergence test required to draw a conclusion.)

*p*-series

*p*-series with . Since , the series converges.

*p*-series:

The symbol means the two series are "asymptotically equivalent"—that is, they either both converge or both diverge because their terms behave so similarly when summed as *n* gets very large. This can be shown by the limit comparison test:

*p*-series with (harmonic series), and so the original series diverges by the limit comparison test.

*p*-series:

*p*-series diverges since (harmonic series), so the larger series diverges by the appropriate direct comparison test.

To see why the inequality holds, consider that when none of the fractions in the square brackets above are actually there; when only 2/2 (which is the same as ) is in the brackets; when only 3/2 (equal to ) and 2/2 (equal to ) are there; when , only 4/2, 3/2, and 2/2 are there; and so forth. Clearly none of these fractions are less than 1 and they never will be, no matter what is used.

The fact that

then implies that

Since the sequence

*n*and are increasing — one may also show that the derivative of the expression is negative for ) and has limit zero (the denominator goes to infinity), so the series converges by the alternating series test.

## Absolute and conditional convergence

editDetermine whether each of the following series converges conditionally, converges absolutely, or diverges. (*Note:* Each "Hint" gives the test or tests that most easily lead to the final conclusion.)

Since this sequence is clearly decreasing to zero, the original series is convergent by the alternating series test. Now, consider the series formed by taking the absolute value of the terms of the original series:

This new series can be compared to a *p*-series:

This sequence has a limit of zero by, for example, L'Hospital's Rule.

That the sequence is decreasing can be verified by, for example, showing that as a continuous function of *x*, its derivative is negative.

This means that the terms definitely decrease starting with the second term ( ). Thus, the series starting at is convergent by the alternating series test; clearly, then, the series starting at also converges (since the two series only differ by one term). Now, consider the series formed by taking the absolute value of the terms of the original series:

This new series of positive terms only can be compared to a *p*-series:

Since the smaller series diverges, the larger one diverges. Alternatively, the integral test can be used to test the convergence of the series of positive terms, since is clearly a continuous, positive function on and, as we have just verified, is also decreasing:

by the substitution ; and this last expression becomes

Since the improper integral diverges, the series of positive terms diverges.

Either way you test it, the series with all positive terms diverges, and this means the original (alternating) series was not absolutely convergent. Thus, the original series is only conditionally convergent.Which implies that

This series of positive terms is asymptotically geometric with :

The equivalence of these series is shown by using a limit comparison test:

Since the limit is positive and finite, and since the simpler series converges because it geometric with (the absolute value of which is less than 1), then the series of positive terms converges by the limit comparison test. Thus the original alternating series is absolutely convergent.

Note, by the way, that a direct comparison test in this case is more difficult (although still possible to do), since

and we need the inequality to go the other way to get a conclusion, since the geometric series converges.

This can be fixed by choosing the new series more carefully:

Because

does not exist (since it continually oscillates within the interval as *n* gets larger)

doesn't exist either. Thus the alternating series test *fails* (it is inconclusive).

However, in such a situation we can use the *divergence test* instead. Since

Note that

Thus

## Taylor series

editFind the Taylor series for the following functions centered at the given values.

All higher derivatives are also zero, so the Taylor expansion of the function:

becomes simply:

The answer is better written in summation form because the Taylor expansion does not terminate. It is convenient to keep the numbers resulting from the power rule separate from the numbers resulting from the chain rule, so we can see the pattern:

The pattern in the derivative values is:

So the Taylor series is:

Thus:

The pattern in the derivative values is tricky:

This is quite an awkward formula for the *n*^{th} derivative. Don't worry at this point about the expression (−1)^{(n+1)/2} in the middle case; we're going to use a simpler version in the final series. Do, however, note the following:

- The
*n*= 0 case [i.e.,*f*(*a*)] doesn't match the rest of the pattern. - When
*n*is positive and even, the value is 0, so the "even-derivative" terms don't contribute anything to the series.

This means we can rewrite Taylor's formula

as

where

Note that the expression 2*n* + 1 is always odd and starts with the value 3 when *n* is 1. Thus *n* = 1 corresponds to the *third* derivative, *n* = 2 the *fifth* derivative, and so on. It should be clear that (−1)^{n+1} gives the correct signs for these derivative values.

Therefore the Taylor series is:

^{2n+1}into 6

^{2n}· 6 and pulling the second 6 out in front of the sum (resulting in a 3 there), this is typically not done to preserve the pattern of having 2

*n*+ 1 everywhere.