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Sequences and Series: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 An impossible report and its nearest repairs

Difficulty: 1 of 3 stars, Stretch

An increasing arithmetic sequence consists entirely of positive integers. A report says that its first four terms sum to $62$ and its next four terms sum to $190$ . Prove that the report is impossible.

Keep $62$ unchanged. What is the smallest absolute change to $190$ that makes the report possible? Find every corrected total achieving that smallest change, and give the corresponding first term and common difference.

Builds on Arithmetic Sequences, Arithmetic Series
1. Hint 1
  If the first term is $a$ and the common difference is $d$ , compare each of terms $5$ through $8$ with the term four places earlier.
2. Hint 2
  The first total gives $2a+3d=31$ . Use integer parity before looking for the nearest possible second total.
Answer

The smallest change is $16$ . The corrected totals are $174$ with $(a,d)=(5,7)$ , and $206$ with $(a,d)=(2,9)$ .

Full solution

Let the first term be the positive integer $a$ and the common difference be the positive integer $d$ .
The first four terms sum to $4a+6d=62$ , or $2a+3d=31$ .
Since $31$ is odd, $d$ must be odd.

Each of terms $5,6,7,8$ is $4d$ more than its partner among terms $1,2,3,4$ .
Their total is therefore $62+16d$ .
A total of $190$ would give $d=(190-62)/16=8$ , contradicting the required odd parity.

The nearest odd integers to $8$ are $7$ and $9$ , each one away.
They give corrected totals $62+16\cdot7=174$ and $62+16\cdot9=206$ , each differing from $190$ by $16$ .
From $a=(31-3d)/2$ , their first terms are $5$ and $2$ , both positive.
Thus both repairs actually give valid increasing positive-integer sequences.
Every other allowable odd $d$ lies farther from $8$ , so no smaller change is possible and these are all nearest repairs.

Answer

The smallest change is $16$ . The corrected totals are $174$ with $(a,d)=(5,7)$ , and $206$ with $(a,d)=(2,9)$ .

Key idea

Aggregate sequence data must satisfy integer constraints; parity can detect an impossible report and certify a nearest repair.
Problem 1: Mark complete
Problem 2 A geometric triple seen through reciprocals

Difficulty: 1 of 3 stars, Stretch

Three consecutive terms of a positive geometric sequence have sum $21$ . Their reciprocals have sum $7/12$ . Find every possible ordered triple of terms, and justify why there are no others.

Builds on Geometric Sequences, Sums and Products of Roots
1. Hint 1
  Write the triple as $a,b,c$ with $b^2=ac$ .
2. Hint 2
  Use the geometric relation to express $1/a+1/b+1/c$ in terms of $a+b+c$ and $b^2$ .
Answer

The ordered triples are $(3,6,12)$ and $(12,6,3)$ .

Full solution

Let the terms be $a,b,c>0$ in order.
Being geometric means $b^2=ac$ .
Consequently $ab+ac+bc=b(a+b+c)$ and $abc=b^3$ .
Dividing gives
$1/a+1/b+1/c=(a+b+c)/b^2.$

The reported sums therefore yield $7/12=21/b^2$ , so $b^2=36$ .
Positivity forces $b=6$ .
The outer terms satisfy $a+c=15$ and $ac=36$ , and hence are the roots of
$z^2-15z+36=(z-3)(z-12).$
Their only possible unordered pair is $\{3,12\}$ .

Both orders are valid: $(3,6,12)$ has ratio $2$ , while $(12,6,3)$ has ratio $1/2$ .
Each sums to $21$ , and each reciprocal sum is
$1/3+1/6+1/12=7/12.$
The sums alone cannot distinguish a positive geometric triple from its reversal, which is why both ordered answers are necessary.

Answer

The ordered triples are $(3,6,12)$ and $(12,6,3)$ .

Key idea

For a geometric triple, express symmetric data through the middle term before solving for the outer terms.
Problem 2: Mark complete
Problem 3 The integer part of a radical telescope

Difficulty: 1 of 3 stars, Stretch

Without decimal approximations to any square root, find the greatest integer not exceeding
$S=\sum_{k=1}^{99}\frac{1}{\sqrt{k}+\sqrt{k+2}}.$
Give an exact expression for $S$ and prove the two integer bounds needed for your answer.

Builds on Rationalizing Denominators
1. Hint 1
  Rationalize a typical denominator. The difference of the radicands is $2$ , so retain the corresponding factor.
2. Hint 2
  This telescope has a gap of two: two terms survive at each end. For the upper bound, compare $\sqrt{101}$ with $9+\sqrt2$ by squaring positive quantities.
Answer

$S=(9+\sqrt{101}-\sqrt2)/2$ and $\lfloor S\rfloor=8$ .

Full solution

Rationalization gives
$1/(\sqrt{k}+\sqrt{k+2})=(\sqrt{k+2}-\sqrt{k})/2.$
Summing from $1$ to $99$ cancels every square root from $\sqrt3$ through $\sqrt{99}$ .
The survivors give $S=(\sqrt{100}+\sqrt{101}-\sqrt1-\sqrt2)/2=(9+\sqrt{101}-\sqrt2)/2$ .

For a lower bound, $\sqrt{101}>10$ and $\sqrt2<2$ , so
$S>(9+10-2)/2=17/2>8.$
For the upper bound, both $\sqrt{101}$ and $9+\sqrt2$ are positive.
Their squares compare as $(9+\sqrt2)^2=83+18\sqrt2>83+18=101$ , since $\sqrt2>1$ .
Thus $\sqrt{101}<9+\sqrt2$ , and substituting gives $S<9$ .

Therefore $8<S<9$ , so the greatest integer not exceeding $S$ is $8$ .
The bounds are exact and do not depend on rounded numerical estimates.

Answer

$S=(9+\sqrt{101}-\sqrt2)/2$ and $\lfloor S\rfloor=8$ .

Key idea

A telescope with a larger gap leaves several boundary terms; exact square comparisons can locate its sum between integers.
Problem 3: Mark complete
Problem 4 Geometric landmarks in an arithmetic sequence

Difficulty: 2 of 3 stars, Challenge

An increasing arithmetic sequence $(a_n)$ has positive terms. The terms $a_2,a_5,a_{11}$ , in that order, form a geometric sequence. The first $20$ arithmetic terms sum to $460$ .

Find $a_n$ explicitly. Then find the position of the next term in the geometric progression that begins $a_2,a_5,a_{11}$ . Prove uniqueness.

Builds on Arithmetic Sequences, Arithmetic Series, Geometric Sequences
1. Hint 1
  Write $a_n=a+(n-1)d$ with $a,d>0$ . The geometric condition is $(a+4d)^2=(a+d)(a+10d)$ .
2. Hint 2
  Cancel the squared terms and factor the remaining relation. After finding $a/d$ , use the arithmetic sum to determine the scale.
Answer

$a_n=2n+2$ . The geometric terms are $6,12,24,48$ , so the next occurs at position $23$ .

Full solution

Let the first term be $a>0$ and common difference be $d>0$ .
The three specified terms are $a+d,a+4d,a+10d$ .
Their geometric condition gives
$(a+4d)^2=(a+d)(a+10d).$
Expanding and canceling yields $6d^2-3ad=0$ , or
$3d(2d-a)=0.$
Since the sequence is increasing, $d\ne0$ , so $a=2d$ .

The sum condition is $460=20(2a+19d)/2=10(23d)$ , hence $d=2$ and $a=4$ .
Thus
$a_n=4+2(n-1)=2n+2.$
These steps uniquely determine both parameters, and the resulting sequence is positive and increasing.

The landmark terms are $a_2=6$ , $a_5=12$ , and $a_{11}=24$ , with common ratio $2$ .
Their next geometric term is $48$ .
Solving $2n+2=48$ gives $n=23$ , a valid positive integer.
Finally the first twenty terms sum to $20(4+42)/2=460$ , confirming every original condition.

Answer

$a_n=2n+2$ . The geometric terms are $6,12,24,48$ , so the next occurs at position $23$ .

Key idea

Special subsequences constrain the shape of a sequence; a separate sum can then determine its scale.
Problem 4: Mark complete
Problem 5 Even-length sums conceal a sign

Difficulty: 2 of 3 stars, Challenge

A geometric sequence has nonzero real first term $a$ and nonzero real common ratio $r$ . The sum of its first two terms is $6$ , and the sum of its first four terms is $30$ . Find all possible sequences.

For each possibility, determine whether the infinite series of reciprocals of its terms converges; if it does, find its sum.

Builds on Geometric Sequences, Geometric Series
1. Hint 1
  Terms three and four together are $r^2$ times the sum of the first two.
2. Hint 2
  Once $r^2$ is known, keep both signs of $r$ . The reciprocal sequence has first term $1/a$ and ratio $1/r$ .
Answer

The possibilities are $a_n=2\cdot2^{n-1}$ and $a_n=-6(-2)^{n-1}$ . Their reciprocal series converge to $1$ and $-1/9$ , respectively.

Full solution

The first two terms give $a(1+r)=6$ , so in particular $r\ne-1$ .
The first four terms can be grouped as
$a(1+r)+ar^2(1+r)=6(1+r^2).$
Equating this to $30$ gives $r^2=4$ , hence $r=2$ or $r=-2$ .

If $r=2$ , then $a=6/3=2$ , producing $2,4,8,16,\ldots$ .
If $r=-2$ , then $a=6/(-1)=-6$ , producing $-6,12,-24,48,\ldots$ .
Their first-two and first-four sums are $6,30$ in both cases, so both are valid.
The equations allow no other choices.

The reciprocals also form geometric sequences.
In the first case their first term is $1/2$ and ratio $1/2$ , yielding sum
$(1/2)/(1-1/2)=1.$
In the second their first term is $-1/6$ and ratio $-1/2$ , yielding sum
$(-1/6)/(1+1/2)=-1/9.$
Both series converge because the absolute value of their ratio is less than $1$ .
The original sequences themselves grow in magnitude; it is their reciprocal series that converge.

Answer

The possibilities are $a_n=2\cdot2^{n-1}$ and $a_n=-6(-2)^{n-1}$ . Their reciprocal series converge to $1$ and $-1/9$ , respectively.

Key idea

Even gaps or blocks reveal an even power of the common ratio and can hide its sign.
Problem 5: Mark complete
Problem 6 Equal partial sums locate a maximum

Difficulty: 2 of 3 stars, Challenge

An arithmetic sequence has partial sums $S_n=a_1+\cdots+a_n$ . You are told that $S_7=S_{19}=133$ . Find the sequence, the greatest value of $S_n$ over all positive integers $n$ , and every $n$ for which $S_n>0$ . Explain how the equal partial sums reveal the location of the maximum.

Builds on Arithmetic Series, Completing the Square, Quadratic Optimization
1. Hint 1
  The sum formula makes $S_n$ a quadratic expression in the index $n$ .
2. Hint 2
  Use $S_{19}-S_7=0$ to relate the first term and common difference. Then complete a square in the resulting formula for $S_n$ .
Answer

$a_n=27-2n$ and $S_n=n(26-n)$ . The maximum is $169$ at $n=13$ only. Positive sums occur exactly for $1\le n\le25$ .

Full solution

Let the first term be $a$ and the common difference be $d$ .
Then
$S_n=n(2a+(n-1)d)/2.$
Subtracting $S_7$ from $S_{19}$ gives $0=(12/2)(2a+25d)$ , so $2a=-25d$ .
Substituting into $S_7=133$ gives $133=(7/2)(-19d)$ , hence $d=-2$ and $a=25$ .

Therefore $a_n=25-2(n-1)=27-2n$ and
$S_n=n(50-2(n-1))/2=n(26-n).$
Completing the square yields
$S_n=169-(n-13)^2.$
This proves that every partial sum is at most $169$ , with equality only at the allowed integer $n=13$ .

For positive integer $n$ , the product $n(26-n)$ is positive exactly when $n<26$ , giving
$1\le n\le25.$
It is zero at $26$ and negative thereafter.
The equal values at $7$ and $19$ are symmetric about their midpoint $13$ , the axis of this quadratic.
The derived square formula proves the symmetry and maximum rather than merely guessing them from those two equal observations.

Answer

$a_n=27-2n$ and $S_n=n(26-n)$ . The maximum is $169$ at $n=13$ only. Positive sums occur exactly for $1\le n\le25$ .

Key idea

Partial sums of an arithmetic sequence form a quadratic in the index, so equal sums can reveal a symmetry axis.
Problem 6: Mark complete
Problem 7 A weighted geometric sum and its exact error

Difficulty: 2 of 3 stars, Challenge

For a positive integer $n$ , define $W_n=\sum_{k=1}^n k/2^k$ . Derive a closed formula for $W_n$ and determine the infinite sum.

Find the least positive integer $n$ for which the difference between the infinite sum and $W_n$ is less than $1/1000$ . Justify the threshold using exact integer comparisons.

Builds on Geometric Series
1. Hint 1
  Write out $W_n$ and $W_n/2$ , aligning equal powers of $2$ before subtracting.
2. Hint 2
  The subtraction leaves a finite geometric sum and one end term. After finding the error, check when $(n+2)/2^n<1/1000$ and prove that this error decreases.
Answer

$W_n=2-(n+2)/2^n$ , the infinite sum is $2$ , and the least $n$ is $14$ .

Full solution

Align $W_n$ with half of itself.
At each denominator $2^k$ for $2\le k\le n$ , the numerator difference is $k-(k-1)=1$ .
The end term from the shifted sum remains, giving
$W_n/2=\sum_{k=1}^n2^{-k}-n/2^{n+1}.$
Since the finite geometric sum is $1-2^{-n}$ , multiplication by $2$ yields
$W_n=2-(n+2)/2^n.$

The error expression $E_n=(n+2)/2^n$ is positive and satisfies $E_{n+1}/E_n=(n+3)/(2n+4)\le2/3$ for $n\ge1$ .
Thus $E_n\le E_1(2/3)^{n-1}$ , which tends to zero as a geometric sequence.
Consequently the partial sums approach $2$ , establishing the infinite sum and showing that $E_n$ is its exact error.

At $n=13$ , $E_{13}=15/8192>1/1000$ , since
$15000>8192.$
At $n=14$ , $E_{14}=16/16384<1/1000$ , since
$16000<16384.$
The positive errors decrease strictly, as their displayed ratio is below $1$ .
Therefore every earlier index also fails, and $14$ is the first success.

Answer

$W_n=2-(n+2)/2^n$ , the infinite sum is $2$ , and the least $n$ is $14$ .

Key idea

Shift and subtract a weighted geometric series to lower its weights; an exact remainder gives a rigorous stopping rule.
Problem 7: Mark complete
Problem 8 A nonlinear sequence with a simple reciprocal sum

Difficulty: 3 of 3 stars, Deep challenge

Let $t>1$ be real. Define $a_1=t$ and $a_{n+1}=a_n^2-a_n+1$ for every positive integer $n$ .

Find a formula for $\sum_{k=1}^N1/a_k$ in terms of $t$ and $a_{N+1}$ , and prove that the infinite series converges. Determine its sum. Finally, find the unique $t$ for which the infinite sum is $3$ .

Builds on Geometric Sequences
1. Hint 1
  Subtract $1$ from the recurrence and factor. Look for a difference involving reciprocals of $a_n-1$ .
2. Hint 2
  Put $b_n=a_n-1>0$ . The rule becomes $b_{n+1}=b_n(1+b_n)$ . Use a fixed geometric lower bound to prove that the leftover reciprocal tends to zero.
Answer

$\sum_{k=1}^N1/a_k=1/(t-1)-1/(a_{N+1}-1)$ . The infinite sum is $1/(t-1)$ ; it equals $3$ exactly when $t=4/3$ .

Full solution

Because $a_1>1$ and $a_{n+1}-1=a_n(a_n-1)$ , every term remains greater than $1$ .
The factorization gives the identity $1/(a_n-1)-1/(a_{n+1}-1)=1/(a_n-1)-1/[a_n(a_n-1)]=1/a_n$ .
Summing it cancels all intermediate reciprocals, leaving
$\sum_{k=1}^N1/a_k=1/(t-1)-1/(a_{N+1}-1).$

To evaluate the infinite sum, it is not enough simply to omit the final term.
Set $b_n=a_n-1$ , so $b_1=t-1>0$ and
$b_{n+1}=b_n(1+b_n).$
This shows $b_{n+1}>b_n$ , hence
$b_n\ge b_1.$
Therefore
$b_{n+1}\ge(1+b_1)b_n=t b_n.$
Repeated application gives
$b_{N+1}\ge(t-1)t^N.$

Consequently $0<1/(a_{N+1}-1)\le1/[(t-1)t^N]$ , and the right side tends to zero because $t>1$ .
The finite-sum formula therefore approaches $1/(t-1)$ , proving convergence and its value.
Requiring this value to be $3$ gives $t-1=1/3$ , so $t=4/3$ , which satisfies the original restriction and is unique.

Answer

$\sum_{k=1}^N1/a_k=1/(t-1)-1/(a_{N+1}-1)$ . The infinite sum is $1/(t-1)$ ; it equals $3$ exactly when $t=4/3$ .

Key idea

A nonlinear recurrence may hide a telescoping reciprocal identity; prove that the surviving boundary term disappears.
Problem 8: Mark complete
Problem 9 Every consecutive-integer representation

Difficulty: 3 of 3 stars, Deep challenge

Find every way to write $315$ as a sum of at least two consecutive positive integers. Give the number of representations and identify the one using the most terms. Prove your search is exhaustive.

Builds on Arithmetic Series
1. Hint 1
  If the first integer is $a$ and the number of terms is $m$ , then $315=m(2a+m-1)/2$ .
2. Hint 2
  Positivity gives $m(m+1)/2\le315$ . Combine this bound with the requirement that $m$ divides $630$ , and check when the resulting $a$ is an integer.
Answer

There are $11$ representations. Their pairs (first integer, number of terms) are $(157,2),(104,3),(61,5),(50,6),(42,7),(31,9),(27,10),(16,14),(14,15),(9,18),(5,21)$ . The longest is $5+6+\cdots+25=315$ .

Full solution

Write a representation as $a+(a+1)+\cdots+(a+m-1)$ with positive integer $a$ and integer $m\ge2$ .
The arithmetic-series formula gives
$m(2a+m-1)=630.$
Hence $m$ must divide $630$ , and
$a=(630/m-m+1)/2.$

Since $a\ge1$ , even the smallest possible sum for a fixed length is
$1+2+\cdots+m=m(m+1)/2.$
Thus
$m(m+1)\le630.$
As $24\cdot25=600$ but $25\cdot26=650$ , we must have
$2\le m\le24.$
Factoring $630=2\cdot3^2\cdot5\cdot7$ gives the complete divisor list in this interval: $2,3,5,6,7,9,10,14,15,18,21$ .

Substituting these lengths into the formula for $a$ , in order, yields $157,104,61,50,42,31,27,16,14,9,5$ .
Every value is a positive integer, so every candidate gives a representation.
Conversely, the bound and divisor condition force every possible length onto this list, proving completeness.
There are eleven entries.
The largest length is $21$ , starting at $5$ and ending at $25$ ; its sum is
$21(5+25)/2=315.$

Answer

There are $11$ representations. Their pairs (first integer, number of terms) are $(157,2),(104,3),(61,5),(50,6),(42,7),(31,9),(27,10),(16,14),(14,15),(9,18),(5,21)$ . The longest is $5+6+\cdots+25=315$ .

Key idea

An arithmetic-series equation can become a bounded divisor search; verify both integrality and positivity for every candidate.
Problem 9: Mark complete
Problem 10 Reconstructing a geometric series from two totals

Difficulty: 3 of 3 stars, Deep challenge

A geometric sequence has a nonzero real first term $a$ and a nonzero real common ratio $r$ . Both its infinite series and the infinite series of the squares of its terms converge. Their sums are $S$ and $Q$ , respectively.

Classify all real pairs $(S,Q)$ that can occur, and for every possible pair recover all choices of $a,r$ . Prove that your conditions are sufficient as well as necessary.

When $S=3$ and $Q=18$ , also find the sum of the absolute values of all the terms.

Builds on Geometric Series, Algebraic Fractions
1. Hint 1
  Convergence requires $|r|<1$ . Write $S=a/(1-r)$ and $Q=a^2/(1-r^2)$ .
2. Hint 2
  Eliminate $a$ to obtain $Q/S^2=(1-r)/(1+r)$ . When you solve for $r$ , check that it is nonzero and lies strictly between $-1$ and $1$ .
Answer

Exactly the pairs with $S\ne0$ , $Q>0$ , and $Q\ne S^2$ occur. Each gives the unique $r=(S^2-Q)/(S^2+Q)$ and $a=2SQ/(S^2+Q)$ . For $S=3,Q=18$ , $a=4$ , $r=-1/3$ , and the sum of absolute values is $6$ .

Full solution

With a nonzero first term, convergence of a geometric series requires $|r|<1$ .
Its sum is $S=a/(1-r)$ , which is nonzero.
The squared terms form a geometric series with ratio $r^2$ , giving
$Q=a^2/(1-r^2)>0.$
Eliminating $a=S(1-r)$ yields
$Q=S^2(1-r)/(1+r).$

Rearranging gives $r=(S^2-Q)/(S^2+Q)$ and then
$a=S(1-r)=2SQ/(S^2+Q).$
Because $r$ is required to be nonzero, we also need $Q\ne S^2$ .
These equations show that if a sequence exists, it is unique.

Conversely, suppose $S\ne0$ , $Q>0$ , and $Q\ne S^2$ .
Define $a,r$ by those formulas.
The denominator is positive and $a\ne0$ .
Also $r\ne0$ by the last restriction.
Since both $S^2$ and $Q$ are positive, $|S^2-Q|<S^2+Q$ , proving $|r|<1$ .
Therefore both series converge.
Substitution into their geometric-sum formulas gives precisely $S$ and $Q$ , proving sufficiency.

For $S=3,Q=18$ , the formulas give $r=-1/3$ and $a=4$ .
Taking absolute values produces a positive geometric series with first term $4$ and ratio $1/3$ , so its sum is
$4/(1-1/3)=6.$
The signed total is smaller because successive terms alternate in sign.

Answer

Exactly the pairs with $S\ne0$ , $Q>0$ , and $Q\ne S^2$ occur. Each gives the unique $r=(S^2-Q)/(S^2+Q)$ and $a=2SQ/(S^2+Q)$ . For $S=3,Q=18$ , $a=4$ , $r=-1/3$ , and the sum of absolute values is $6$ .

Key idea

Two series totals can determine a hidden sequence; reverse the algebra and verify convergence to establish an exact existence classification.
Problem 10: Mark complete