0
$\begingroup$

Determine if the series converges or diverges. If it converges what does it converge to?

${\sum_{n=1}^\infty \frac{4}{n(n+3)}}$

Here's what I have been able to figure out

By the limit comparison test ${\sum\frac{1}{x^2}}$

${\lim_{x\to\infty}\frac{4n^2}{n^2 + 3n} = 4\Rightarrow}$ Convergence

But I can't figure out how to tell what it converges to, it's not an alternating series, or a geometric series that I can tell. And we haven't yet covered taylor series. It might be a power series, but I don't see any x value to tell what it converges to.

$\endgroup$
2

3 Answers 3

3
$\begingroup$

\begin{align*} &\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n}-\dfrac{1}{n+3}\right)\\ &=\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)+\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right) \end{align*}

$\endgroup$
0
$\begingroup$

If $a$ is a positive integer then, if $m > a$,

$\begin{array}\\ s(a, m) &=\sum_{n=1}^{m} \dfrac1{n(n+a)}\\ &=\sum_{n=1}^{m} \dfrac1{a}(\dfrac1{n}-\dfrac1{n+a})\\ &=\dfrac1{a}\sum_{n=1}^{m} (\dfrac1{n}-\dfrac1{n+a})\\ &=\dfrac1{a}\sum_{n=1}^{m} \dfrac1{n}-\dfrac1{a}\sum_{n=1}^{m} \dfrac1{n+a}\\ &=\dfrac1{a}\sum_{n=1}^{m} \dfrac1{n}-\dfrac1{a}\sum_{n=a+1}^{a+m} \dfrac1{n}\\ &=\dfrac1{a}\left(\sum_{n=1}^{m} \dfrac1{n}-\sum_{n=a+1}^{a+m} \dfrac1{n}\right)\\ &=\dfrac1{a}\left(\sum_{n=1}^{a} \dfrac1{n}+\sum_{n=a+1}^{m} \dfrac1{n}-\sum_{n=a+1}^{m} \dfrac1{n}-\sum_{n=m+1}^{a+m} \dfrac1{n}\right)\\ &=\dfrac1{a}\left(\sum_{n=1}^{a} \dfrac1{n}-\sum_{n=m+1}^{a+m} \dfrac1{n}\right)\\ \end{array} $

so

$\begin{array}\\ s(a, m)-\dfrac1{a}\sum_{n=1}^{a} \dfrac1{n} &=-\dfrac1{a}\sum_{n=m+1}^{a+m} \dfrac1{n}\\ \text{so that}\\ |s(a, m)-\dfrac1{a}\sum_{n=1}^{a} \dfrac1{n}| &=\dfrac1{a}\sum_{n=m+1}^{a+m} \dfrac1{n}\\ &\le\dfrac1{a}\sum_{n=m+1}^{a+m} \dfrac1{m+1}\\ &=\dfrac1{a}\dfrac{a}{m+1}\\ &=\dfrac1{m+1}\\ \end{array} $

Therefore $\lim_{m \to \infty} s(a, m) =\dfrac1{a}\sum_{n=1}^{a} \dfrac1{n} $.

$\endgroup$
0
$\begingroup$

Observe that it can be showed that, in general (it's quite evident why this happens): $$\sum_{i = 1}^{\infty} \left(\frac{1}{i} - \frac{1}{(i+p)} \right) = \sum_{i = 1}^{p} \frac{1}{i}$$ But see that: $$\sum_{n = 1}^{\infty} \frac{4}{(n(n+3))} = 4 \sum_{n = 1}^{\infty} \frac{1}{3n} - \frac{1}{3(n+3)} = \sum_{n = 1}^{\infty} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right)$$ In this way, $p = 3$, then we have: $$\sum_{n = 1}^{\infty} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right) = \sum_{n = 1}^{3} \frac{4}{3}\left(\frac{1}{n} - \frac{1}{n+3} \right) = 73/45$$ Logo a sequência converge.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .