Prove that the series $\sum_{n=1}^{\infty}\frac{2}{n(n+2)}$ is convergent

Question

Prove that the series $\sum_{n=1}^{\infty}\frac{2}{n(n+2)}$ is convergent

I know the series is convergent if the sequence of partial sums is convergent.

$\begin{align*} s_n&=a_1+a_2+\cdots +a_n\\ &=\frac{2}{1\cdot 3}+\frac{2}{2\cdot 4}+\frac{2}{3\cdot 5}+\cdots+\frac{2}{n(n+2)}\\ &<\frac{1}{2 \cdot 1}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n-1)}\\ &=(1- \frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots+(\frac{1}{n-1}+\frac{1}{n})\\ &=1-\frac{1}{n}\\ &<1 \end{align*}$

Thus we see that $1$ is an upper bound for the sequence of partial sums, and clearly the sequence is increasing since $0\leq a_n\leq a_{n+1}$. Therefore by the monotone convergence theorem we know that the series $\sum_{n=1}^{\infty}\frac{2}{n(n+2)}$ is convergent.

Answer 1

Hint: $$\dfrac{2}{n(n+2)} = \dfrac{2}{n^2 +2n} \leq \dfrac{2}{n^2}$$ for all $n \geq 1$, and $$\sum_{n=1}^{\infty}\dfrac{2}{n^2}$$ is proportional to a convergent $p$-series. These are both non-negative series. Now apply Limit Comparison.

Answer 2

I don't know where your first inequality comes from. You can solve your problem as follows:\begin{align}\frac2{1.3}+\frac2{2.4}+\frac2{3.5}+\cdots&<\frac2{1.2}+\frac2{2.3}+\frac2{3.4}+\cdots\\&=2\left(1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots\right)\\&<2.\end{align}

Answer 3

Hint:

$$\frac2{n(n+2)}=\frac1n-\frac1{n+2}$$

Answer 4

Another hint: $$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2},$$ So that the sum telescopes.

Specifically $(1-1/3)+(1/2-1/4)+(1/3-1/5) \cdots =3/2.$