Calculus/Sequences and Series/Exercises

compare adjacent terms algebraically or take a derivative

monotonically decreasing (strictly decreasing starting with the second term)

One may either consider how ${\displaystyle a_{n+1}}$ compares to ${\displaystyle a_{n}}$ algebraically and try to show that one is greater than the other, or take the derivative of ${\displaystyle a_{n}}$ with respect to ${\displaystyle n}$ and check where it is positive or negative.

Algebraically, since

${\displaystyle a_{n+1}-a_{n}={\frac {n+1}{2^{n+1}}}-{\frac {n}{2^{n}}}={\frac {n+1}{2^{n+1}}}-{\frac {2n}{2^{n+1}}}={\frac {1-n}{2^{n+1}}},}$

we see that ${\displaystyle a_{n+1}<a_{n}}$ for ${\displaystyle n>1}$. 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 ${\displaystyle n=1}$ and ${\displaystyle n=2}$:

${\displaystyle a_{1}=1/2}$

${\displaystyle a_{2}=2/4=1/2}$

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

Using calculus,

${\displaystyle {\frac {d}{dn}}\,{\frac {n}{2^{n}}}={\frac {2^{n}\cdot 1-n\cdot 2^{n}\ln 2}{(2^{n})^{2}}}={\frac {2^{n}(1-n\ln 2)}{2^{2n}}}={\frac {1-n\ln 2}{2^{n}}},}$

which is negative for ${\displaystyle n>1/\ln 2\approx 1.44>1}$. The rest of the argument is the same as before.

consider what kind of values are taken on by the numerator and denominator, and use the previous answer

bounded from below and from above (both)

The sequence is bounded from below because the terms are clearly positive (greater than 0) for all values of ${\displaystyle n}$. Also, since the sequence is decreasing (see the previous problem), the maximum value of the sequence must be the value of the first term. So the sequence is bounded from above (by the value 1/2), as well.

use the previous two answers to make a conclusion, or take a limit

converges

By the previous two answers, the sequence is bounded from below and monotonically decreasing, thus by a theorem it must converge.

To show this directly, consider the limit

${\displaystyle \lim _{n\to \infty }{\frac {n}{2^{n}}}=\lim _{n\to \infty }{\frac {1}{2^{n}\ln 2}}=0.}$

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.

take a limit

converges to 2

The series converges to 2 since

${\displaystyle s=\lim _{n\to \infty }s_{n}=\lim _{n\to \infty }\left(2-{\frac {1}{3^{n}}}\right)=2.}$

${\displaystyle a_{n}=s_{n}-s_{n-1}=\left(2-{\frac {1}{3^{n}}}\right)-\left(2-{\frac {1}{3^{n-1}}}\right)={\frac {1}{3^{n-1}}}-{\frac {1}{3^{n}}}={\frac {3}{3^{n}}}-{\frac {1}{3^{n}}}={\frac {2}{3^{n}}}}$

Note that the series turns out to be geometric, since

${\displaystyle a_{n}=2\,(1/3)^{n}=a\,r^{n}.}$

sum of an infinite geometric series

4

The series is

${\displaystyle \sum _{n=0}^{\infty }3\left({\frac {1}{4}}\right)^{n}}$

and so is geometric with first term ${\displaystyle a=3}$ and common ratio ${\displaystyle r=1/4}$. So

${\displaystyle s={\frac {a}{1-r}}={\frac {3}{1-1/4}}=4.}$

sum of an infinite geometric series

telescoping series

1

Note that

${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{n^{2}-n}}=\sum _{n=2}^{\infty }{\frac {1}{n(n-1)}}=\sum _{n=2}^{\infty }\left({\frac {1}{n-1}}-{\frac {1}{n}}\right)}$

by partial fractions. So

${\displaystyle s=\lim _{N\to \infty }s_{N}=\lim _{N\to \infty }\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\left({\frac {1}{3}}-{\frac {1}{4}}\right)+\ldots +\left({\frac {1}{N-1}}-{\frac {1}{N}}\right).}$

All but the first and last terms cancel out, so

${\displaystyle s=\lim _{N\to \infty }\left(1-{\frac {1}{N}}\right)=1.}$

rewrite so that all exponents are n

−1/5

The series simplifies to

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}2^{n}}{3^{n}\cdot 2}}=\sum _{n=1}^{\infty }{\frac {1}{2}}\left({\frac {-2}{3}}\right)^{n}}$

and so is geometric with common ratio ${\displaystyle r=-2/3}$ and first term ${\displaystyle -1/3}$. Thus

${\displaystyle s={\frac {-1/3}{1-(-2/3)}}=-1/5.}$

p-series

converges

This is a p-series with ${\displaystyle p=2}$. Since ${\displaystyle p>1}$, the series converges.

geometric series

converges

This is a geometric series with common ratio ${\displaystyle r=1/2}$, and so converges since ${\displaystyle |r|<1}$.

limit comparison test

diverges

This series can be compared to a p-series:

${\displaystyle \sum _{n=1}^{\infty }{\frac {n}{n^{2}+1}}\sim \sum _{n=1}^{\infty }{\frac {n}{n^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{n}}}$

The ${\displaystyle \sim }$ 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:

${\displaystyle \lim _{n\to \infty }\left({\frac {n}{n^{2}+1}}\div {\frac {1}{n}}\right)=\lim _{n\to \infty }\left({\frac {n}{n^{2}+1}}\cdot {\frac {n}{1}}\right)=\lim _{n\to \infty }{\frac {n^{2}}{n^{2}+1}}=1}$

Since the limit is positive and finite, the two series either both converge or both diverge. The simpler series diverges because it is a p-series with ${\displaystyle p=1}$ (harmonic series), and so the original series diverges by the limit comparison test.

direct comparison test

diverges

This series can be compared to a smaller p-series:

${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{\ln n}}\geq \sum _{n=2}^{\infty }{\frac {1}{n}}}$

The p-series diverges since ${\displaystyle p=1}$ (harmonic series), so the larger series diverges by the appropriate direct comparison test.

divergence test

diverges

The terms of this series do not have a limit of zero. Note that when ${\displaystyle n>1}$,

${\displaystyle {\frac {n!}{2^{n}}}={\frac {n}{2}}\cdot \left[{\frac {n-1}{2}}\cdot {\frac {n-2}{2}}\dots {\frac {2}{2}}\right]\cdot {\frac {1}{2}}\geq {\frac {n}{2}}\cdot (1)\cdot {\frac {1}{2}}={\frac {n}{4}}}$

To see why the inequality holds, consider that when ${\displaystyle n=2}$ none of the fractions in the square brackets above are actually there; when ${\displaystyle n=3}$ only 2/2 (which is the same as ${\displaystyle [n-1]/2}$) is in the brackets; when ${\displaystyle n=4}$ only 3/2 (equal to ${\displaystyle [n-1]/2}$) and 2/2 (equal to ${\displaystyle [n-2]/2}$) are there; when ${\displaystyle n=5}$, 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 ${\displaystyle n>1}$ is used.

The fact that

${\displaystyle \lim _{n\to \infty }{\frac {n}{4}}=\infty }$

then implies that

${\displaystyle \lim _{n\to \infty }{\frac {n!}{2^{n}}}=\infty }$

Therefore the series diverges by the divergence test.

alternating series test

converges

This is an alternating series:

${\displaystyle \sum _{n=1}^{\infty }{\frac {\cos \pi n}{n}}=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}}$

Since the sequence

${\displaystyle |a_{n}|={\frac {1}{n}}}$

decreases to 0, the series converges by the alternating series test.

alternating series test

converges

Since the terms alternate, consider the sequence

${\displaystyle |a_{n}|={\frac {1}{n\ln n-1}}}$

This sequence is clearly decreasing (since both n and ${\displaystyle \ln n}$ are increasing — one may also show that the derivative of the expression is negative for ${\displaystyle n\geq 2}$) and has limit zero (the denominator goes to infinity), so the series converges by the alternating series test.

alternating series test and either direct comparison test or integral test

converges conditionally

This series alternates, so consider the sequence

${\displaystyle |a_{n}|={\frac {1}{\sqrt {n}}}}$

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:

${\displaystyle \sum |a_{n}|=\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}}$

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

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}\geq \sum _{n=1}^{\infty }{\frac {1}{n}}}$

Since the smaller series diverges, the larger one diverges. But this means the original (alternating) series was not absolutely convergent. (This last fact can also be shown using an integral test.) Therefore, the original series is only conditionally convergent.

alternating series test and either integral test or direct comparison test

converges conditionally

This series alternates, so consider the sequence

${\displaystyle |a_{n}|={\frac {\ln n}{n}}}$

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

${\displaystyle \lim _{n\to \infty }{\frac {\ln n}{n}}=\lim _{n\to \infty }{\frac {1/n}{1}}=0}$

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

${\displaystyle {\frac {d}{dx}}\,{\frac {\ln x}{x}}={\frac {1-\ln x}{x^{2}}}<0{\mbox{ if }}x>e}$

This means that the terms definitely decrease starting with the second term (${\displaystyle n=3}$). Thus, the series starting at ${\displaystyle n=3}$ is convergent by the alternating series test; clearly, then, the series starting at ${\displaystyle n=2}$ 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:

${\displaystyle \sum |a_{n}|=\sum _{n=2}^{\infty }{\frac {\ln n}{n}}}$

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

${\displaystyle \sum _{n=2}^{\infty }{\frac {\ln n}{n}}\geq \sum _{n=2}^{\infty }{\frac {1}{n}}}$

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 ${\displaystyle f(x)={\frac {\ln x}{x}}}$ is clearly a continuous, positive function on ${\displaystyle [2,\infty )}$ and, as we have just verified, is also decreasing:

${\displaystyle \int _{2}^{\infty }{\frac {\ln x}{x}}\,dx=\lim _{t\to \infty }\int _{2}^{t}{\frac {\ln x}{x}}\,dx=\lim _{t\to \infty }\int _{\ln 2}^{\ln t}u\,du}$

by the substitution ${\displaystyle u=\ln x}$; and this last expression becomes

${\displaystyle \lim _{t\to \infty }{\Bigl (}{\frac {1}{2}}u^{2}{\bigr |}_{\ln 2}^{\ln t}{\Bigr )}=\lim _{t\to \infty }{\frac {1}{2}}(\ln t)^{2}-{\frac {1}{2}}(\ln 2)^{2}=\infty }$

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.

divergence test

diverges

This series is alternating, but note that by L'Hospital's Rule

${\displaystyle \lim _{n\to \infty }{\frac {n}{(\ln n)^{2}}}=\lim _{n\to \infty }{\frac {1}{2(\ln n)(1/n)}}=\lim _{n\to \infty }{\frac {n}{2\ln n}}}$

${\displaystyle {\mbox{ }}=\lim _{n\to \infty }{\frac {1}{2/n}}=\lim _{n\to \infty }{\frac {n}{2}}=\infty }$

Which implies that

${\displaystyle \lim _{n\to \infty }{\frac {(-1)^{n}n}{(\ln n)^{2}}}}$

does not exist, and hence by the divergence test, the series diverges.

limit comparison test with geometric series

converges absolutely

While this alternating series can be shown to converge by the alternating series test, it can also be shown that the absolute value of the terms form a convergent series, and this is sufficient to conclude absolute convergence of the original series. Thus we will skip the former test and show only the latter.

${\displaystyle \sum _{n=1}^{\infty }|a_{n}|=\sum _{n=1}^{\infty }{\frac {2^{n}}{e^{n}-1}}}$

This series of positive terms is asymptotically geometric with ${\displaystyle r=2/e}$:

${\displaystyle \sum _{n=1}^{\infty }{\frac {2^{n}}{e^{n}-1}}\sim \sum _{n=1}^{\infty }(2/e)^{n}}$

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

${\displaystyle \lim _{n\to \infty }\left[{\frac {2^{n}}{e^{n}-1}}\div (2/e)^{n}\right]=\lim _{n\to \infty }\left({\frac {2^{n}}{2^{n}}}\cdot {\frac {e^{n}}{e^{n}-1}}\right)=\lim _{n\to \infty }{\frac {1}{1-1/e^{n}}}=1}$

Since the limit is positive and finite, and since the simpler series converges because it geometric with ${\displaystyle r=2/e}$ (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

${\displaystyle \sum _{n=1}^{\infty }{\frac {2^{n}}{e^{n}-1}}\geq \sum _{n=1}^{\infty }(2/e)^{n}}$

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:

${\displaystyle \sum _{n=1}^{\infty }{\frac {2^{n}}{e^{n}-1}}\leq \sum _{n=1}^{\infty }{\frac {2^{n}}{e^{n}-.5e^{n}}}=\sum _{n=1}^{\infty }{\frac {2^{n}}{.5e^{n}}}=\sum _{n=1}^{\infty }2(2/e)^{n}}$

Comparing the original series to the new (convergent geometric) series gives the desired result.

divergence test

diverges

Since this is an alternating series, we can try the alternating series test. Consider the absolute value of the terms:

${\displaystyle |a_{n}|={\frac {1}{\sin ^{2}n}}}$

Because

${\displaystyle \lim _{n\to \infty }\sin ^{2}n}$

does not exist (since it continually oscillates within the interval ${\displaystyle [0,1]}$ as n gets larger)

${\displaystyle \lim _{n\to \infty }{\frac {1}{\sin ^{2}n}}}$

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

${\displaystyle \lim _{n\to \infty }{\frac {(-1)^{n}}{\sin ^{2}n}}}$

also does not exist (and thus the terms of the series do not converge to 0), the original series diverges by the divergence test.

ratio test

converges absolutely

Because of the factorials in this series, we try the ratio test:

${\displaystyle \lim _{n\to \infty }\left|{\frac {(-1)^{n+1}(n+1)!}{[2(n+1)]!}}\div {\frac {(-1)^{n}n!}{(2n)!}}\right|=\lim _{n\to \infty }{\frac {(n+1)!}{n!}}\cdot {\frac {(2n)!}{(2n+2)!}}=\lim _{n\to \infty }{\frac {(n+1)}{(2n+2)(2n+1)}}=\lim _{n\to \infty }{\frac {1}{2(2n+1)}}=0}$

Since the limit is less than 1, the series converges absolutely by the ratio test.

divergence test

diverges

Although this is an alternating series, neither the numerator nor the denominator have infinite limits, so it is likely that a divergence test will work.

Note that

${\displaystyle \lim _{n\to \infty }{\frac {e^{1/n}}{\arctan n}}={\frac {e^{0}}{\pi /2}}={\frac {2}{\pi }}\neq 0}$

Thus

${\displaystyle \lim _{n\to \infty }{\frac {(-1)^{n}e^{1/n}}{\arctan n}}\neq 0}$

In fact, the latter limit does not exist. So, by the divergence test, the series diverges.

don't try to use the summation formula—the Taylor series has a finite number of terms

This is a polynomial, so the Taylor series expansion terminates. Repeatedly differentiating and plugging in 1, we get:

${\displaystyle f(a)=16}$

${\displaystyle f'(x)=2(1+3x)(3)=6(1+3x)\longrightarrow f'(a)=24}$

${\displaystyle f''(x)=6(3)=18\longrightarrow f''(a)=18}$

${\displaystyle f'''(x)=0\longrightarrow f'''(a)=0}$

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

${\displaystyle f(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\dots }$

becomes simply:

${\displaystyle f(x)=16+24(x-1)+{\frac {18}{2}}(x-1)^{2}}$

${\displaystyle =16+24(x-1)+9(x-1)^{2}}$

Note that the answer is not simplified any further (e.g., multiplying everything out and combining like terms), because the resulting representation would no longer appear to be centered at 1.

avoid using the quotient rule

This rational function can be written in a form that is easier to repeatedly differentiate:

${\displaystyle f(x)=(5-2x)^{-1}}$

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:

${\displaystyle f(a)=1}$

${\displaystyle f'(x)=(-1)(5-2x)^{-2}(-2)=(1)(5-2x)^{-2}(2)\longrightarrow f'(a)=1\cdot 2}$

${\displaystyle f''(x)=(-1)(-2)(5-2x)^{-3}(-2)(-2)=(1\cdot 2)(5-2x)^{-3}(2^{2})\longrightarrow f''(a)=(1\cdot 2)(2^{2})}$

${\displaystyle f'''(x)=(-1)(-2)(-3)(5-2x)^{-4}(-2)(-2)(-2)=(1\cdot 2\cdot 3)(5-2x)^{-4}(2^{3})\longrightarrow f'''(a)=(1\cdot 2\cdot 3)(2^{3})}$

${\displaystyle f^{(4)}(x)=(-1)(-2)(-3)(-4)(5-2x)^{-4}(-2)(-2)(-2)(-2)=(1\cdot 2\cdot 3\cdot 4)(5-2x)^{-4}(2^{4})\longrightarrow f^{(4)}(a)=(1\cdot 2\cdot 3\cdot 4)(2^{4})}$

The pattern in the derivative values is:

${\displaystyle f^{(n)}(a)=n!\,2^{n}}$

So the Taylor series is:

${\displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}=\sum _{n=0}^{\infty }{\frac {n!\,2^{n}}{n!}}(x-2)^{n}}$

${\displaystyle =\sum _{n=0}^{\infty }2^{n}(x-2)^{n}}$

rewrite so the trigonometric function is not squared

This function can be rewritten to make it easier to repeatedly differentiate. By the half-angle trigonometric identity for sine:

${\displaystyle f(x)={\frac {1}{2}}-{\frac {1}{2}}\cos 6x}$

Thus:

${\displaystyle f(a)=(1/2)-(1/2)\cos(3\pi /2)=1/2}$

${\displaystyle f'(x)=(1/2)(\sin 6x)(6)\longrightarrow f'(a)=(1/2)(-1)(6)}$

${\displaystyle f''(x)=(1/2)(\cos 6x)(6^{2})\longrightarrow f''(a)=0}$

${\displaystyle f'''(x)=(1/2)(-\sin 6x)(6^{3})\longrightarrow f'''(a)=(1/2)(1)(6^{3})}$

${\displaystyle f^{(4)}(x)=(1/2)(-\cos 6x)(6^{4})\longrightarrow f'''(a)=0}$

${\displaystyle f^{(5)}(x)=(1/2)(\sin 6x)(6^{5})\longrightarrow f'''(a)=(1/2)(-1)(6^{5})}$

The pattern in the derivative values is tricky:

${\displaystyle f^{(n)}(a)={\begin{cases}1/2&{\text{if }}n=0\\(1/2)(-1)^{(n+1)/2}(6^{n})&{\text{if }}n>0{\text{ and odd}}\\0&{\text{if }}n>0{\text{ and even}}\end{cases}}}$

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:

1. The n = 0 case [i.e., f(a)] doesn't match the rest of the pattern.
2. 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

${\displaystyle f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}}$

as

${\displaystyle f(x)=f(a)+\sum _{n=1}^{\infty }{\frac {f^{(2n+1)}(a)}{(2n+1)!}}(x-a)^{2n+1}}$

where

${\displaystyle f^{(2n+1)}(a)=(1/2)(-1)^{n+1}(6^{2n+1})}$

Note that the expression 2n + 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:

${\displaystyle f(x)=1/2+\sum _{n=1}^{\infty }{\frac {(1/2)(-1)^{n+1}(6^{2n+1})}{(2n+1)!}}(x-\pi /4)^{2n+1}}$

${\displaystyle ={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}\,6^{2n+1}}{(2n+1)!}}{\Bigl (}x-{\frac {\pi }{4}}{\Bigr )}^{2n+1}}$

While this could be simplified a bit more by splitting 6²ⁿ⁺¹ into 6²ⁿ · 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 2n + 1 everywhere.