Investigate the convergence of $ \sum_{n=1}^{\infty } (-1)^{n}\frac{n+2}{n(n+1)} $

Question

I am supposed to investigate the convergence of $ \sum_{n=1}^{\infty } (-1)^{n}\frac{n+2}{n(n+1)} $. I'm unsure whether to use Leibniz' criterion or a comparison test and I really can't start. Thanks

Answer 1

You can directly use Leibniz' criterion, but you have to show the absolute value of the general term is non-increasing. Here, I suggest an (arguably) simpler way to see what's happening and prove convergence:

You have $$ \frac{n+2}{n(n+1)} = \frac{1}{n+1} + \frac{2}{n(n+1)} $$ and therefore $$ \sum_{n=1}^\infty (-1)^n \frac{n+2}{n(n+1)} = \sum_{n=1}^\infty \frac{(-1)^n}{n+1} + 2\sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)} $$ The first series on the right-hand-side converges conditionally by Leibniz's criterion, the second converges absolutely.

Answer 2

Hint:

$$ {{n + 2} \over {n\left( {n + 1} \right)}} - {{n + 3} \over {\left( {n + 1} \right)\left( {n + 2} \right)}} = {{n + 4} \over {n\left( {n + 1} \right)\left( {n + 2} \right)}} $$

Answer 3

Just to show the longer, more formulaic approach:

Let $a_n: = \frac{n+2}{n(n+1)}$. Then $$a_{n+1} - a_n = \frac{n+3}{(n+1)(n+2)} - \frac{n+2}{n(n+1)}$$ Using a common denominator we have $$a_{n+1}-a_n = \frac{n^2 + 3n - n^2 - 4n - 4}{n(n+1)(n+2)} = \frac{-(n+4)}{n(n+1)(n+2)}$$ and then clearly $$a_{n+1}-a_n = \frac{-(n+4)}{n(n+1)(n+2)} < 0 \hspace{2mm} \forall n \in \mathbb{N}$$

So $a_n$ is a decreasing sequence. Furthermore, it is clear that $a_n \rightarrow 0$ as $n \rightarrow \infty$, so by the alternating series test (or Leibniz criterion), the sum $\sum_{n=1}^{\infty} (-1)^n a_n$ converges.

Answer 4

1) $a_n = \dfrac{n+2}{n(n+1)}= \dfrac{1}{n} +\dfrac{1}{n(n+1)}$.

$a_{n+1}=\dfrac{1}{n+1}+ \dfrac{1}{(n+1)(n+2)};$

Hence

$a_n > a_{n+1}$ (Why?).

2) $\lim_{ n \rightarrow \infty}a_n=$

$\lim_{n \rightarrow \infty}(\dfrac{1}{n}+\dfrac{1}{n(n+1)})=0$(why?).

Apply Leibniz criterion.