Question

Determine if the series converges or diverges. If the series converges, find its sum.
9
Σ n(n+3)
n=1


OA. The series diverges.
OB. The series converges to
11
2
7
OC. The series converges to
2
D. The series converges
15
-
2

Answer 1

The true statement about the series $\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}$ is that (a) the series diverges

How to determine if the series diverges or converges?

The series is given as:

$\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}$

Take the limit of the function to infinity

$\lim_{n \to \infty} \frac{9}{n(n +3)}$

This gives

$\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * (\infty +3)}$

Evaluate the sum

$\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * \infty}$

Evaluate the product

$\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty}$

Evaluate the quotient

$\lim_{n \to \infty} \frac{9}{n(n +3)} = 0$

Since the limit is 0, then it means that the series diverges

Hence, the true statement about the series $\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}$ is that (a) the series diverges

Read more about series at:

https://brainly.com/question/337693

#SPJ1

Answer 2

Answer:

B. The series converges to $\displaystyle{\frac{11}{2}}$.

Step-by-step explanation:

Before evaluating the infinite series, the expression can be decomposed as the sum of two fractions (partial fraction decomposition) as follows.

Let $\textit{A}$ and $\textit{B}$ be constants such that

                                       ${\displaystyle{\frac{9}{n\left(n+3\right)}}}} \ \ = \ \ \displaystyle{\frac{A}{n} \ \ \ + \ \ \frac{B}{n+3}}$

Multiply both sides of the equation by the denominator of the left fraction,

$n\left(n+3\right)$, yielding

                                              $9 \ \ = \ \ A\left(n+3\right) \ \ + \ \ B \-\hspace{0.045cm} n$

Now, let $n \ = \ 0$, thus

                                             $\-\hspace{0.2cm} 9 \ \ = \ \ A\left(0 + 3\right) \ + \ B\left(0\right) \\ \\ 3 \-\hspace{0.035cm} A \ = \ \ 9 \\ \\ \-\hspace{0.11cm} A \ \ = \ \ 3$.

Likewise, let $n \ = \ -3$, then

                                            $\-\hspace{0.5cm} 9 \ \ = \ \ A\left(-3 + 3\right) \ + \ B\left(-3\right) \\ \\ -3 \-\hspace{0.035cm} B \ = \ \ 9 \\ \\ \-\hspace{0.44cm} B \ \ = \ \ -3$

Hence,

                                       $\displaystyle{\sum_{n=1}^{\infty} {\frac{9}{n\left(n+3\right)}}} \ = \ \displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right)$.

First and foremost, write the nth partial sum (first nth terms) of the series,

          $\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ - \frac{3}{4} + \ \frac{3}{2} \ - \frac{3}{5} \ + \frac{3}{3} \ - \frac{3}{6} \ + \frac{3}{4} \ - \frac{3}{7}} \\ \\ \\ \-\hspace{3.58cm} + \ \displaystyle{\frac{3}{5} \ - \ \frac{3}{8} \ + \ \frac{3}{6} \ - \ \frac{3}{9} \ + \ \frac{3}{7} \ - \ \frac{3}{10}} \\ \\ \\ \-\hspace{3.58cm} + \ \ \dots$

                                                $+ \ \ \displaystyle{\frac{3}{n-3} \ - \ \frac{3}{n} \ + \ \frac{3}{n-2} \ - \ \frac{3}{n+1}} \\ \\ \\ \ + \ \frac{3}{n-1} \ - \ \frac{3}{n+2} \ + \ \frac{3}{n} - \ \frac{3}{n+3}}$.

Notice that the expression forms a telescoping sum where subsequent terms cancel each other, leaving only

        $\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ + \ \frac{3}{2} \ + \frac{3}{3} \ - \ \frac{3}{n+1}} - \ \frac{3}{n+2} \ - \ \frac{3}{n+3}}}$.

To determine if this infinite series converges or diverges, evaluate the limit of the nth partial sum as $n \ \rightarrow \ \infty$,

         $\displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \lim_{n \to \infty} \left(\displaytstyle{\frac{11}{2} \ - \ \frac{3}{n+1} \ - \ \frac{3}{n+2} \ + \ - \ \frac{3}{n+3}\right) \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2} \ - \ 0 \ - \ 0 \ - \ 0} \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2}$