consider the following:
$\sum_{n=1}^{\infty} \frac{19}{(2n+1)(2n-1)}$
(a) find the first four terms: $a_1, a_2, a_3,$ and $a_4$
(b) find the first four terms of the sequence of the partial sums, that is, $s_1, s_2, s_3,$ and $s_4$
(c) use the results of part (b) to find a formula for $s_n$
(d) determine whether the given series is convergent or divergent.
show all your work clearly.

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Find $a_1$
Substitute $n = 1$ into $\frac{19}{(2n + 1)(2n - 1)}$.
$a_1=\frac{19}{(2(1)+1)(2(1)-1)}=\frac{19}{3\times1}=\frac{19}{3}$
## Step2: Find $a_2$
Substitute $n = 2$ into $\frac{19}{(2n + 1)(2n - 1)}$.
$a_2=\frac{19}{(2(2)+1)(2(2)-1)}=\frac{19}{5\times3}=\frac{19}{15}$
## Step3: Find $a_3$
Substitute $n = 3$ into $\frac{19}{(2n + 1)(2n - 1)}$.
$a_3=\frac{19}{(2(3)+1)(2(3)-1)}=\frac{19}{7\times5}=\frac{19}{35}$
## Step4: Find $a_4$
Substitute $n = 4$ into $\frac{19}{(2n + 1)(2n - 1)}$.
$a_4=\frac{19}{(2(4)+1)(2(4)-1)}=\frac{19}{9\times7}=\frac{19}{63}$
# Answer:
$a_1=\frac{19}{3}$, $a_2=\frac{19}{15}$, $a_3=\frac{19}{35}$, $a_4=\frac{19}{63}$

### Part (b)
# Explanation:
## Step1: Find $S_1$
$S_1=a_1=\frac{19}{3}$
## Step2: Find $S_2$
$S_2=a_1 + a_2=\frac{19}{3}+\frac{19}{15}=\frac{95 + 19}{15}=\frac{114}{15}=\frac{38}{5}$
## Step3: Find $S_3$
$S_3=S_2 + a_3=\frac{38}{5}+\frac{19}{35}=\frac{266+19}{35}=\frac{285}{35}=\frac{57}{7}$
## Step4: Find $S_4$
$S_4=S_3 + a_4=\frac{57}{7}+\frac{19}{63}=\frac{477 + 19}{63}=\frac{496}{63}$ (Wait, correction: $\frac{57}{7}=\frac{513}{63}$, so $\frac{513+19}{63}=\frac{532}{63}=\frac{76}{9}$)
Wait, let's recalculate $S_2$: $\frac{19}{3}+\frac{19}{15}=\frac{95 + 19}{15}=\frac{114}{15}=\frac{38}{5}$ (correct). $S_3=\frac{38}{5}+\frac{19}{35}=\frac{38\times7+19}{35}=\frac{266 + 19}{35}=\frac{285}{35}=\frac{57}{7}$ (correct). $S_4=\frac{57}{7}+\frac{19}{63}=\frac{57\times9+19}{63}=\frac{513+19}{63}=\frac{532}{63}=\frac{76}{9}$ (yes, 532 ÷7=76, 63÷7=9)
# Answer:
$S_1=\frac{19}{3}$, $S_2=\frac{38}{5}$, $S_3=\frac{57}{7}$, $S_4=\frac{76}{9}$

### Part (c)
# Explanation:
## Step1: Analyze the pattern of $S_n$
We have $S_1=\frac{19}{3}=\frac{19(2\times1 - 1)}{2(2\times1+1)}$? Wait, $S_1=\frac{19}{3}=\frac{19(1)}{3}$, $S_2=\frac{38}{5}=\frac{19\times2}{5}$, $S_3=\frac{57}{7}=\frac{19\times3}{7}$, $S_4=\frac{76}{9}=\frac{19\times4}{9}$
So the numerator is $19n$, denominator is $2n + 1$? Wait, $S_1:\frac{19\times1}{2\times1+1}=\frac{19}{3}$ (matches), $S_2:\frac{19\times2}{2\times2+1}=\frac{38}{5}$ (matches), $S_3:\frac{19\times3}{2\times3+1}=\frac{57}{7}$ (matches), $S_4:\frac{19\times4}{2\times4+1}=\frac{76}{9}$ (matches)
So the formula for $S_n$ is $\frac{19n}{2n + 1}$
# Answer:
$S_n=\frac{19n}{2n + 1}$

### Part (d)
# Explanation:
## Step1: Find the limit of $S_n$ as $n\to\infty$
We have $S_n=\frac{19n}{2n + 1}$. Divide numerator and denominator by $n$: $\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{19}{2+\frac{1}{n}}$
## Step2: Evaluate the limit
As $n\to\infty$, $\frac{1}{n}\to0$, so $\lim_{n\to\infty}S_n=\frac{19}{2+0}=\frac{19}{2}$
Since the limit of the partial sums exists (is finite), the series is convergent.
# Answer:
The series is convergent, and its sum is $\frac{19}{2}$

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Find \(a_1\)

Step2: Find \(a_2\)

Step3: Find \(a_3\)

Step4: Find \(a_4\)

Step1: Find \(S_1\)

Step2: Find \(S_2\)

Step3: Find \(S_3\)

Step4: Find \(S_4\)

Step1: Analyze the pattern of \(S_n\)

Answer:

Part (b)