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Consider the series ∑n=1∞5n2+n.
The general formula for the sum of the first n terms is Sn=
. Your answer should be in terms of n.
The sum of a series is defined as the limit of the sequence of partial sums, which means
∑n=1∞5n2+n=limn→∞(
)=
.
Select all true statements (there may be more than one correct answer):
A. Most of the terms in each partial sum cancel out.
B. The series converges.
C. The series is a p-series.
D. The series is a telescoping series (i.e., it is like a collapsible telescope).
E. The series is a geometric series.

Consider the series n15n2n The general formula for the sum of the first n terms is Sn Your answer should be in terms of n The sum of a series is defined as the class=

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LammettHash

(a) Decompose the summand into partial fractions:

[tex]\dfrac5{n^2+n} = \dfrac5{n(n+1)} = \dfrac an+\dfrac b{n+1}[/tex]

[tex]\implies 5=a(n+1)+bn=(a+b)n+a[/tex]

[tex]\implies a+b=0\text{ and }a=5 \implies b=-5[/tex]

[tex]\implies\displaystyle\sum_{n=1}^\infty\frac5{n^2+n} = 5\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)[/tex]

The n-th partial sum for the series is

[tex]S_n = 5\displaystyle\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)[/tex]

which can be simplified significantly by examinging consective terms in the sum:

[tex]\displaystyle S_n = 5\left(1-\frac12\right) + 5\left(\frac12-\frac13\right) + 5\left(\frac13-\frac14\right) + \cdots + 5\left(\frac1{n-1}-\frac1n\right) + 5\left(\frac1n-\frac1{n+1}\right)[/tex]

[tex]\implies S_n = \boxed{5\left(1-\dfrac1{n+1}\right)}[/tex]

(b) Using the result of (a), you then get

[tex]\displaystyle\sum_{n=1}^\infty\frac5{n^2+n} = \lim_{n\to\infty}\boxed{5\left(1-\frac1{n+1}\right)} = \boxed{5}[/tex]

(c) As shown in (a), the partial sum is simplified because of the reasons given in options A and D, and the result of (b) says that B is also correct.

Space

Answer:

Part a.  [tex]\displaystyle S_n = 5 - \frac{5}{n + 1}[/tex]

Part b.  [tex]\displaystyle \lim_{n \to \infty} (5 - \frac{5}{n + 1}) = 5[/tex]

Part c. A, B, and D

General Formulas and Concepts:

Algebra I

  • Terms/Coefficients
  • Factoring

Pre-Calculus

  • Partial Fraction Decomposition

Calculus

Limits

  • Limit Rule [Variable Direct Substitution]:                                                    [tex]\displaystyle \lim_{x \to c} x = c[/tex]
  • Limit Property [Addition/Subtraction]:                                                         [tex]\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)[/tex]

Sequences

Series

  • Definition of a convergent or divergent series

Telescoping Series:                                                                                             [tex]\displaystyle \sum^\infty_{n = 1} (b_n - b_{n + 1}) = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + ... + (b_n - b_{n + 1}) + ...[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n}[/tex]

Step 2: Rewrite Sum

  1. Factor:                                                                                                           [tex]\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \sum^\infty_{n = 1} \frac{5}{n(n + 1)}[/tex]
  2. Break up [Partial Fraction Decomposition]:                                                 [tex]\displaystyle \frac{5}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}[/tex]
  3. Simplify [Common Denominator]:                                                                 [tex]\displaystyle 5 = A(n + 1) + Bn[/tex]
  4. [Decomp] Substitute in n = 0:                                                                       [tex]\displaystyle 5 = A(0 + 1) + B(0)[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle 5 = A[/tex]
  6. [Decomp] Substitute in n = -1:                                                                       [tex]\displaystyle 5 = A(-1 + 1) + B(-1)[/tex]
  7. Simplify:                                                                                                         [tex]\displaystyle 5 = -B[/tex]
  8. Solve:                                                                                                             [tex]\displaystyle B = -5[/tex]
  9. [Decomp] Substitute in variables:                                                                 [tex]\displaystyle \frac{5}{n(n + 1)} = \frac{5}{n} + \frac{-5}{n + 1}[/tex]
  10. Simplify:                                                                                                         [tex]\displaystyle \frac{5}{n(n + 1)} = \frac{5}{n} - \frac{5}{n + 1}[/tex]
  11. Substitute in decomp [Sum]:                                                                         [tex]\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \sum^\infty_{n = 1} \bigg( \frac{5}{n} - \frac{5}{n + 1} \bigg)[/tex]

Step 3: Find Sum

  1. Find Sₙ terms:                                                                                                    [tex]\displaystyle \sum^\infty_{n = 1} \bigg( \frac{5}{n} - \frac{5}{n + 1} \bigg) = (5 - \frac{5}{2}) + (\frac{5}{2} - \frac{5}{3}) + (\frac{5}{3} - \frac{5}{4}) + (\frac{5}{4} - 1) + ... + ( \frac{5}{n} - \frac{5}{n + 1}) + ...[/tex]
  2. Find general Sₙ formula:                                                                               [tex]\displaystyle S_n = 5 - \frac{5}{n + 1}[/tex]
  3. Find Sum [Take limit]:                                                                                   [tex]\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = \lim_{n \to \infty} S_n[/tex]
  4. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           [tex]\displaystyle \displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = 5 + 0[/tex]
  5. Simplify:                                                                                                         [tex]\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n} = 5[/tex]

∴ the sum converges by the Telescoping Series.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Convergence Tests (BC Only)

Book: College Calculus 10e

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