Question

Consider the series ∑n=1∞5n2+n.

Answer 1

(a) Decompose the summand into partial fractions:

$\dfrac5{n^2+n} = \dfrac5{n(n+1)} = \dfrac an+\dfrac b{n+1}$

$\implies 5=a(n+1)+bn=(a+b)n+a$

$\implies a+b=0\text{ and }a=5 \implies b=-5$

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

The n-th partial sum for the series is

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

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

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

$\implies S_n = \boxed{5\left(1-\dfrac1{n+1}\right)}$

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

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

(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.

Answer 2

Answer:

Part a.  $\displaystyle S_n = 5 - \frac{5}{n + 1}$

Part b.  $\displaystyle \lim_{n \to \infty} (5 - \frac{5}{n + 1}) = 5$

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]:                                                    $\displaystyle \lim_{x \to c} x = c$
• Limit Property [Addition/Subtraction]:                                                         $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Sequences

Series

• Definition of a convergent or divergent series

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \sum^\infty_{n = 1} \frac{5}{n^2 + n}$

Step 2: Rewrite Sum

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

Step 3: Find Sum

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

∴ 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