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Consider the rational-exponential expression fn(x)=∑k=1n(x2k(k+1)+3x2k)f_n(x) = \sum_{k=1}^{n} \left( \frac{x^2}{k(k+1)} + \frac{3x}{2^k} \right). By recognizing telescoping and geometric summations to combine like powers of xx, find the limiting simplified expression F(x)=lim⁡n→∞fn(x)F(x) = \lim_{n \to \infty} f_n(x) as the number of terms approaches infinity.

Answer and steps

Final answer

x2+3xx^2 + 3x

Steps

01

Separate the summation by powers of xx

To evaluate the limiting expression F(x)=lim⁡n→∞fn(x)F(x) = \lim_{n \to \infty} f_n(x), we can separate the terms containing like powers of xx and evaluate each summation individually.

The terms inside the summation depend on the index kk, while xx acts as a constant with respect to the sum. Distributing the summation yields:

fn(x)=x2∑k=1n1k(k+1)+3x∑k=1n12kf_n(x) = x^2 \sum_{k=1}^{n} \frac{1}{k(k+1)} + 3x \sum_{k=1}^{n} \frac{1}{2^k}

Taking the limit as n→∞n \to \infty:

F(x)=x2(∑k=1∞1k(k+1))+3x(∑k=1∞12k)F(x) = x^2 \left( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \right) + 3x \left( \sum_{k=1}^{\infty} \frac{1}{2^k} \right)

02

Evaluate the telescoping series for the x2x^2 coefficient

Using partial fraction decomposition, rewrite the general term:

1k(k+1)=1k−1k+1\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

Writing out the first few terms of the partial sum:

∑k=1n(1k−1k+1)=(1−12)+(12−13)+⋯+(1n−1n+1)\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)

Notice that intermediate terms cancel out in pairs:

∑k=1n1k(k+1)=1−1n+1\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}

Taking the limit as n→∞n \to \infty:

lim⁡n→∞(1−1n+1)=1−0=1\lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1

Thus, the coefficient of x2x^2 is 11, contributing:

x2⋅1=x2x^2 \cdot 1 = x^2

03

Evaluate the geometric series for the xx coefficient

The second sum is an infinite geometric series:

∑k=1∞12k=12+14+18+…\sum_{k=1}^{\infty} \frac{1}{2^k} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots

Here, the first term is a=12a = \frac{1}{2} and the common ratio is r=12r = \frac{1}{2}. Since ∣r∣<1|r| < 1, the sum converges according to the infinite geometric series formula S=a1−rS = \frac{a}{1 - r}:

S=121−12=1212=1S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1

Thus, the coefficient of 3x3x is 11, contributing:

3x⋅1=3x3x \cdot 1 = 3x

04

Combine the simplified terms

Adding the simplified components together gives the limiting expression:

F(x)=x2+3xF(x) = x^2 + 3x

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misapplies the infinite geometric series formula by using 1 instead of the first term 1/2 in the numerator, resulting in 2 instead of 1 for the geometric sum.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    F(x)=x2∑k=1∞1k(k+1)+3x∑k=1∞12kF(x) = x^2 \sum_{k=1}^{\infty} \frac{1}{k(k+1)} + 3x \sum_{k=1}^{\infty} \frac{1}{2^k}

  2. Step 2

    ∑k=1∞(1k−1k+1)=1\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1

  3. Step 3

    ∑k=1∞(12)k=11−12=2\sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k = \frac{1}{1 - \frac{1}{2}} = 2

    Correct

    ∑k=1∞(12)k=121−12=1\sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1

    The sum starts at k=1k = 1, so the first term is a=12a = \frac{1}{2}, not 11. The infinite sum formula is S=a1−r=1/21−1/2=1S = \frac{a}{1 - r} = \frac{1/2}{1 - 1/2} = 1.

  4. Step 4

    F(x)=x2(1)+3x(2)  ⟹  F(x)=x2+6xF(x) = x^2(1) + 3x(2) \implies F(x) = x^2 + 6x

    Correct

    F(x)=x2(1)+3x(1)=x2+3xF(x) = x^2(1) + 3x(1) = x^2 + 3x

    Your final combination followed logically from your geometric sum value, but it inherited the error from Step 3.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly factored out powers of xx from the summation index.

You used partial fractions accurately to evaluate the telescoping series.

You substituted your calculated sum values back into the expression systematically.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Geometric series index misalignment

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You applied the geometric series formula 11−r\frac{1}{1 - r}, which assumes the series begins at k=0k = 0, instead of the general formula S=a1−rS = \frac{a}{1 - r} where aa is the actual first term at k=1k = 1.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the initial index

  • For any geometric series ∑k=k0∞rk\sum_{k=k_0}^{\infty} r^k, the sum is a1−r\frac{a}{1-r}, where a=rk0a = r^{k_0}.

  • Write out the first term explicitly before applying the summation formula.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Identify the first term aa for k=1k = 1: a=(12)1=12a = \left( \frac{1}{2} \right)^1 = \frac{1}{2}.

  2. Identify the common ratio r=12r = \frac{1}{2}.

  3. Evaluate the sum: S=a1−r=121−12=1212=1S = \frac{a}{1 - r} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1.

  4. Substitute both series values back into F(x)F(x): F(x)=x2(1)+3x(1)=x2+3xF(x) = x^2(1) + 3x(1) = x^2 + 3x.

Final answer

x2+3xx^2 + 3x

NEXT STEPS
Final Takeaways from Your Teacher

  • Before applying a1−r\frac{a}{1-r}, always write down the explicit numerical value of the very first term aa.

  • Watch out for summation bounds: note whether the index starts at k=0k=0 or k=1k=1.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly decomposes 1/(k(k+1)) as 1/k + 1/(k+1) instead of subtracting, failing to recognize that telescoping cancellation requires a difference of terms.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    1k(k+1)=1k+1k+1\frac{1}{k(k+1)} = \frac{1}{k} + \frac{1}{k+1}

    Correct

    1k(k+1)=1k−1k+1\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

    When performing partial fraction decomposition, the numerators must subtract so that the terms in the numerator cancel: 1k−1k+1=(k+1)−kk(k+1)=1k(k+1)\frac{1}{k} - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)}. Adding them yields 2k+1k(k+1)\frac{2k+1}{k(k+1)}, not 1k(k+1)\frac{1}{k(k+1)}.

  2. Step 2

    ∑k=1∞1k(k+1) diverges\sum_{k=1}^{\infty} \frac{1}{k(k+1)} \text{ diverges}

    Correct

    ∑k=1∞1k(k+1)=lim⁡n→∞∑k=1n(1k−1k+1)=lim⁡n→∞(1−1n+1)=1\sum_{k=1}^{\infty} \frac{1}{k(k+1)} = \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1

    Because of the addition sign in your decomposition, you viewed the sum as a sum of divergent harmonic-type terms rather than a telescoping series where adjacent terms cancel to leave 11.

  3. Step 3

    ∑k=1∞12k=1/21−1/2=1\sum_{k=1}^{\infty} \frac{1}{2^k} = \frac{1/2}{1 - 1/2} = 1

  4. Step 4

    F(x)  does not exist because the harmonic sum diverges\text{ does not exist because the harmonic sum diverges}

    Correct

    F(x) = x^2(1) + 3x(1) = x^2 + 3x

    Since the first sum actually converges to 11, both coefficients exist and combine to form F(x)=x2+3xF(x) = x^2 + 3x.

WHAT YOU DID WELL
Compliments from Your Teacher

You separated the expression into distinct summations corresponding to powers of xx.

You correctly calculated the sum of the infinite geometric series ∑k=1∞12k=1\sum_{k=1}^{\infty} \frac{1}{2^k} = 1 using the formula S=a1−rS = \frac{a}{1-r}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Partial fraction sign error

You wrote a plus sign instead of a minus sign between the partial fractions 1k\frac{1}{k} and 1k+1\frac{1}{k+1}.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You did not verify the partial fraction decomposition by recombining the terms over a common denominator, overlooking that telescoping cancellation requires a difference of consecutive terms.

REASONING PATH
Your ACEpath teacher’s suggested approach
Recombine to verify decomposition

  • Always test partial fractions: Ak+Bk+1=A(k+1)+Bkk(k+1)\frac{A}{k} + \frac{B}{k+1} = \frac{A(k+1) + Bk}{k(k+1)}. For the numerator to be constant 11, the variable terms must cancel, requiring A+B=0A + B = 0.

  • Check your decomposition mentally: 1k−1k+1=(k+1)−kk(k+1)=1k(k+1)\frac{1}{k} - \frac{1}{k+1} = \frac{(k+1)-k}{k(k+1)} = \frac{1}{k(k+1)}.

Recognize telescoping structure

  • A series telescopes when its general term can be written in the form ak−ak+1a_k - a_{k+1}, allowing all intermediate terms to cancel out.

  • Write out the first two partial sum terms to ensure opposite signs appear and cancel.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Correct the partial fraction decomposition: 1k(k+1)=1k−1k+1\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

  2. Expand the partial sum for the x2x^2 coefficient: ∑k=1n(1k−1k+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{n+1}.

  3. Take the limit as n→∞n \to \infty: lim⁡n→∞(1−1n+1)=1\lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1.

  4. Combine the terms with your correct geometric series value: F(x)=x2(1)+3x(1)=x2+3xF(x) = x^2(1) + 3x(1) = x^2 + 3x.

Final answer

x^2 + 3x

NEXT STEPS
Final Takeaways from Your Teacher

  • Review partial fraction decomposition by checking that A(k+1)+Bk=1A(k+1) + Bk = 1 gives A=1,B=−1A = 1, B = -1.

  • Practice expanding telescoping series by writing the first two and the last term of partial sums.

  • Solve the targeted practice exercises below to master combining series limits.

OverviewACEpath Teacher's Diagnosis

03 The student treats the upper index limit of the telescoping sum as terminating at 0 without accounting for the boundary term 1/(n+1), but incorrectly shifts the remaining starting term to 1/2.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    1k(k+1)=1k−1k+1  ⟹  ∑k=1n(1k−1k+1)=(1−12)+(12−13)+…\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \implies \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots

  2. Step 2

    Remaining first term taken as 12 instead of 1  ⟹  ∑k=1∞1k(k+1)=12\text{Remaining first term taken as } \frac{1}{2} \text{ instead of } 1 \implies \sum_{k=1}^{\infty} \frac{1}{k(k+1)} = \frac{1}{2}

    Correct

    ∑k=1n(1k−1k+1)=1−1n+1  ⟹  lim⁡n→∞(1−1n+1)=1−0=1\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1} \implies \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1

    In the expansion (1−12)+(12−13)+…(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots, the term −12-\frac{1}{2} is cancelled by +12+\frac{1}{2}. The uncancelled first boundary term from k=1k=1 is 11, not 12\frac{1}{2}. Therefore, the sum evaluates to 11 rather than 12\frac{1}{2}.

  3. Step 3

    ∑k=1∞(12)k=121−12=1\sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1

  4. Step 4

    F(x) = 12\frac{1}{2}x^2 + 3x

    Correct

    F(x)=x2∑k=1∞1k(k+1)+3x∑k=1∞12k=x2(1)+3x(1)=x2+3xF(x) = x^2 \sum_{k=1}^{\infty} \frac{1}{k(k+1)} + 3x \sum_{k=1}^{\infty} \frac{1}{2^k} = x^2(1) + 3x(1) = x^2 + 3x

    Your linear combination followed logically from your computed coefficients, but the final expression carries over the coefficient error from Step 2.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly used partial fractions to express the rational term as a telescoping difference.

You evaluated the infinite geometric series accurately using the sum formula.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Telescoping boundary term error

You took 12\frac{1}{2} instead of 11 as the surviving boundary term of the telescoping sum.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusion regarding which components of the initial pair cancel out, mistakenly treating the subtracted term −12-\frac{1}{2} as the surviving initial value rather than recognizing that 11 is never cancelled.

REASONING PATH
Your ACEpath teacher’s suggested approach
Pair Cancellation Tracking

  • In ∑k=1n(ak−ak+1)\sum_{k=1}^n (a_k - a_{k+1}), intermediate adjacent terms cancel in opposite-sign pairs: −a2-a_2 cancels with +a2+a_2.

  • Always circle or underline the very first positive term a1a_1 and cross out the pairs that follow to see what remains uncancelled.

Closed-Form Partial Sum Check

  • The finite partial sum of ∑k=1n(1k−1k+1)\sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right) equals 1−1n+11 - \frac{1}{n+1}.

  • Write out the explicit nn-th partial sum formula SnS_n before taking the limit as n→∞n \to \infty.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Write the finite partial sum explicitly: Sn=∑k=1n(1k−1k+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1S_n = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1}.

  2. Evaluate the infinite limit: lim⁡n→∞Sn=1−0=1\lim_{n \to \infty} S_n = 1 - 0 = 1.

  3. Multiply by the respective powers of xx: F(x)=x2(1)+3x(1)=x2+3xF(x) = x^2(1) + 3x(1) = x^2 + 3x.

Final answer

x^2 + 3x

NEXT STEPS
Final Takeaways from Your Teacher

  • When expanding telescoping series, draw explicit cancellation slashes between consecutive terms to clearly identify which outer boundary terms remain.

  • Always express the partial sum SnS_n in terms of nn before letting n→∞n \to \infty to avoid losing track of boundary values.

Lock it in

Question

What is the sum of an infinite geometric series starting at index k=1k = 1 with first term aa and common ratio ∣r∣<1|r| < 1?

Answer

S=a1−rS = \frac{a}{1 - r}, where a=r1a = r^1 (the first term). The formula 11−r\frac{1}{1-r} only works when the first term is 1 (i.e., starting at k=0k = 0).

Similar question

01

Evaluate the infinite geometric series: S=∑k=1∞53kS = \sum_{k=1}^{\infty} \frac{5}{3^k}.

Answer

S=52S = \frac{5}{2}. The first term is a=53a = \frac{5}{3} and common ratio is r=13r = \frac{1}{3}. Using S=a1−rS = \frac{a}{1-r}, we get S=5/31−1/3=5/32/3=52S = \frac{5/3}{1 - 1/3} = \frac{5/3}{2/3} = \frac{5}{2}.

02

Find the limiting expression G(x)=lim⁡n→∞∑k=1n(2xk(k+2)+4x23k)G(x) = \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{2x}{k(k+2)} + \frac{4x^2}{3^k} \right) by evaluating the telescoping and geometric parts.

Answer

G(x)=2x2+32xG(x) = 2x^2 + \frac{3}{2}x. For the x2x^2 coefficient: ∑k=1∞43k=4/31−1/3=2\sum_{k=1}^{\infty} \frac{4}{3^k} = \frac{4/3}{1 - 1/3} = 2. For the xx coefficient: 2k(k+2)=1k−1k+2\frac{2}{k(k+2)} = \frac{1}{k} - \frac{1}{k+2}, whose sum telescopes to 1+12=321 + \frac{1}{2} = \frac{3}{2}. Combining terms gives 2x2+32x2x^2 + \frac{3}{2}x.

03

Determine the value of cc such that lim⁡n→∞∑k=2n(cx2k+x2k2−1)=34x2+3x\lim_{n \to \infty} \sum_{k=2}^{n} \left( \frac{c x}{2^k} + \frac{x^2}{k^2 - 1} \right) = \frac{3}{4}x^2 + 3x.

Answer

c=12c = 12. For the x2x^2 coefficient: ∑k=2∞1(k−1)(k+1)=12∑k=2∞(1k−1−1k+1)=12(1+12)=34\sum_{k=2}^{\infty} \frac{1}{(k-1)(k+1)} = \frac{1}{2} \sum_{k=2}^{\infty} \left(\frac{1}{k-1} - \frac{1}{k+1}\right) = \frac{1}{2}\left(1 + \frac{1}{2}\right) = \frac{3}{4}, which matches. For the xx coefficient: ∑k=2∞c2k=c⋅1/41−1/2=c⋅12\sum_{k=2}^{\infty} \frac{c}{2^k} = c \cdot \frac{1/4}{1 - 1/2} = c \cdot \frac{1}{2}. Setting c2=3\frac{c}{2} = 3 yields c=12c = 12.

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