Seen Questions

Convergence of a sequence

Monotone convergence theorem

Sequences and series

(a)

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence of non-negative real numbers. Which of the following ensures that $\{a_n\}_{n=0}^{\infty}$ converges? Select all that apply.

• $\{a_n\}_{n=0}^{\infty}$ is bounded below and non-decreasing.
• There exists $L\in\mathbb R$ such that for each $\varepsilon>0$ there corresponds $K\in\mathbb N$ such that $|a_n-L|<\varepsilon$ whenever $n>K$.
• $\sup\{a_n:n\in\mathbb N\}$ exists.
• $\{a_n\}_{n=0}^{\infty}$ is bounded above and non-increasing.
• $\inf\{a_n:n\in\mathbb N\}$ exists.

(b)

Consider the sequence $\{a_n\}_{n=0}^{\infty}$ given by

Given that this sequence is decreasing and converges, find

(c)

Let $\{c_n\}_{n=0}^{\infty}$ and $\{d_n\}_{n=0}^{\infty}$ be sequences of strictly positive real numbers satisfying

For $n\ge1$, define

Determine which statements are true:

• $\displaystyle\sum_{n=1}^{\infty}b_n$ is convergent by the integral test.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ is divergent by the ratio test.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ is divergent by the root test.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ is convergent by the ratio test.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ is convergent by the root test.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ is divergent by the integral test.

(d)

Let $\{a_n\}_{n=0}^{\infty}$ and $\{b_n\}_{n=0}^{\infty}$ be sequences such that

and

where $\alpha$ is a positive integer. Since these formulas describe only the tail, the associated series may be started at $n=1$; finitely many initial terms do not affect convergence.

Determine which of the following statements hold:

• $\displaystyle\sum_{n=1}^{\infty}b_n$ converges if and only if, $\displaystyle\sum_{n=1}^{\infty}a_n$ diverges.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ converges or diverges independently of whether $\displaystyle\sum_{n=1}^{\infty}a_n$ converges or not.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ converges if and only if, $\displaystyle\sum_{n=1}^{\infty}a_n$ converges.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ diverges if and only if, $\displaystyle\sum_{n=1}^{\infty}a_n$ converges.
• $\displaystyle\sum_{n=1}^{\infty}b_n$ diverges if and only if, $\displaystyle\sum_{n=1}^{\infty}a_n$ diverges.

Solution

(a)

The second statement is exactly the $\varepsilon$-$N$ definition of convergence, so it is sufficient.

The fourth statement is also sufficient. Being bounded above gives no lower bound, but the terms are non-negative, so $a_n\ge0$ supplies one and the Monotone Convergence Theorem gives convergence.

The other three statements are not sufficient:

• a non-decreasing sequence that is only bounded below can be $a_n=n$;
• a finite supremum does not prevent oscillation, as in $a_n=0,1,0,1,\ldots$;
• a finite infimum also does not prevent the same oscillation.

Therefore the correct selections are

(b)

Because the sequence converges and the square-root function is continuous on its domain, taking limits in the recurrence gives

Before squaring, observe that

so any admissible limit must satisfy $L\le8$. Squaring now gives

and hence

The algebraic candidates are $L=2$ and $L=13$, but $L=13$ violates the restriction $L\le8$ introduced before squaring. The remaining value checks in the original recurrence:

Therefore

(c)

For $n\ge1$, the $n$th root of $b_n$ is

because $c_n,d_n>0$. Therefore

The root test proves convergence. The $n=0$ term, if included, is just one finite initial term and has no effect on convergence. The given information does not determine the ratio-test limit or provide the smoothness and monotonicity needed for the integral test. Thus the only correct statement is

(d)

The error terms are smaller than the displayed leading terms:

Consequently,

so the terms are eventually positive. Their ratio satisfies

Since this limit is finite and strictly positive, the Limit Comparison Test gives

If $\alpha=1$, both tails behave like the harmonic series and diverge; if $\alpha>1$, both behave like a convergent $p$-series. The correct selections are the third and fifth listed statements. The third is

Taking logical complements gives the fifth:

The sequence parts separate the definition of convergence from the sufficient conditions of monotonicity and boundedness. The series parts use the root test for exponential-type terms and limit comparison for matching power laws. In both, only the tail matters.

Taylor polynomial

Bound a Taylor approximation error

Taylor polynomials and a Lagrange remainder

Aaron and Cindy are investigating a function $f:\mathbb R\to\mathbb R$ and its first eight Taylor polynomials about $x_0=0.3$:

Aaron compares the first six polynomials and records these features near $x_0$: $p_1$ is decreasing at $x_0$; $p_2=p_1$; and, just to the right of $x_0$, $p_3-p_2$ and $p_4-p_3$ are positive, while $p_5-p_4$ is negative. These observations replace the plots, which are not included here.

(a)

(i)

Select all derivatives of $f$ at $0.3$ that are zero:

(ii)

Select all derivatives of $f$ at $0.3$ that are positive.

(iii)

Select all derivatives of $f$ at $0.3$ that are negative.

(b)

Aaron tells Cindy that the coefficient of $(x-0.3)^6$ in $p_6(x)$ is

and the coefficient of $(x-0.3)^7$ in $p_7(x)$ is

They both agree that the graph of $f^{(6)}$ looks like a straight line and decide to assume this is true.

They consider $x_1=0.5$ and want to use Lagrange's form of the remainder to find a bound on

(i)

Is knowing the value of $a_7$ necessary to calculate the error bound found using Lagrange's form of the remainder? Select all true statements:

• The value of $a_7$ is necessary for these particular values of $a_6$ and $x_1$.
• The value of $a_7$ is not necessary for these particular values of $a_6$ and $x_1$.
• The value of $a_7$ would be necessary, regardless of the values of $a_6$ and $x_1$.
• The value of $a_7$ would be necessary for some values of $a_6$ and $x_1$ and unnecessary for others.

(ii)

Find the smallest error bound guaranteed by Lagrange's form of the remainder:

(c)

The fact that the graph in part (b) looks like a straight line with non-zero slope suggests that $f$ is a polynomial of degree $7$. Discuss this conclusion, making clear what follows from an exact global assumption and what a graph alone can establish.

Solution

For a Taylor polynomial about $x_0=0.3$,

The coefficient of $(x-0.3)^k$ is therefore $f^{(k)}(0.3)/k!$. Since $k!>0$, the sign of that coefficient is the sign of the corresponding derivative.

(a)

Consecutive Taylor polynomials differ by a single term:

The polynomial $p_1$ is decreasing at $x_0$, so

Because $p_2=p_1$, its quadratic coefficient vanishes, so

For $x>x_0$, the factor $(x-x_0)^k$ is positive. The recorded signs of $p_3-p_2$, $p_4-p_3$ and $p_5-p_4$ therefore give

Thus the selections are:

• zero: ${f''(0.3)}$;
• positive: ${f^{(3)}(0.3),f^{(4)}(0.3)}$;
• negative: ${f'(0.3),f^{(5)}(0.3)}$.

(b)(i)

The Lagrange remainder for $p_5$ involves $f^{(6)}(c)$ for some $c$ between $0.3$ and $0.5$. The value $a_6$ gives $f^{(6)}(0.3)$, but the assumed straight-line graph of $f^{(6)}$ also needs its slope to determine the largest absolute value on the interval. That slope is determined by $a_7$.

For these particular values, the first statement is true:

The fourth statement is also true in general: if $x_1=x_0$, the error is zero without knowing $a_7$, whereas over a nontrivial interval the slope can change the maximum of $|f^{(6)}|$. The second and third statements are false.

(b)(ii)

The Taylor coefficients give

and

Because $f^{(6)}$ is assumed to be a straight line,

On the interval $[0.3,0.5]$,

A linear function takes its extreme values on a closed interval at the endpoints. Therefore

Lagrange's form of the remainder for the degree-five polynomial is

for some $c\in(0.3,0.5)$. Consequently,

(c)

If $f^{(6)}$ is assumed to be exactly a straight line with non-zero slope on all of $\mathbb R$, then $f^{(7)}$ is a non-zero constant and $f^{(8)}=0$. Integrating this identity repeatedly shows that $f$ is a polynomial of degree exactly seven:

A graph that merely looks linear, or is known to be linear only on the interval used for the error bound, does not establish this global conclusion; it supports the assumption locally but cannot determine $f$ outside that interval.

The plots encode derivative signs through Taylor coefficients. The sixth and seventh coefficients turn the straight-line assumption for $f^{(6)}$ into a bound on its maximum, which is the quantity Lagrange's remainder needs.

Convergence has a central interval

Root test

Alternating-series test

A power series and its endpoints

It is “play with series” night and Chelsea and Finn ask you to consider the power series

(a)

Let

Which of the following statements is true?

• $a_n>0$ for all $n\ge1$;
• $a_n$ takes both positive and negative values at an infinite number of positive integers $n$;
• $a_n>0$ for all $n>1$;
• $a_n>0$ if $n$ is odd and $a_n<0$ if $n$ is even.

(b)

Chelsea asks you to use an appropriate criterion to find the open interval of convergence $I=(a,b)$ for the power series (1) and to enter the values of $a$ and $b$.

(c)

Determine what happens at the endpoints $x=a$ and $x=b$. Select the true statements:

• The power series (1) at $x=b$ is convergent.
• The power series (1) at $x=a$ is not convergent.
• The power series (1) at $x=b$ is not convergent.
• The power series (1) at $x=a$ is convergent.

In a few sentences, justify your answers, explaining your reasoning.

Solution

For $n=1$,

If $n$ is even, then

If $n$ is odd and $n>1$, then

Hence the correct statement in part (a) is

For $n>1$,

Both bounds have $n$th root tending to $1$, so

Apply the root test to the general term of (1):

The series converges absolutely when

Thus the open interval is

The root test does not decide the endpoints, so they must be substituted into the original series.

At $x=b=13$, the factor $(x-7)/6$ is $1$ and the series becomes

For $n\ge4$,

The comparison series $\sum 1/\sqrt n$ diverges, so the series at $x=b$ is not convergent.

At $x=a=1$, the factor $(x-7)/6$ is $-1$ and the series becomes

The first series converges by the alternating-series test, while the second is the divergent harmonic series. Therefore the series at $x=a$ is also not convergent.

Thus the endpoint statements are

The radius depends only on the $n$th-root limit of the slowly varying coefficients, which is $1$. The endpoints need separate arguments: at the left endpoint the alternating $1/\sqrt n$ part converges but the harmonic part does not and at the right endpoint the positive $1/\sqrt n$ behaviour forces divergence.