Question

Show that the following series converges :
summation (infinity , n= 1) [ 1/(2n+3)(2n+5)]?

Answer 1

To prove that the series

$$\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{(2n + 3)(2n + 5)}}}$$

converges, let's compare it with another series

$$\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^2}}}}$$

We can see that for any $n \ge 1$

$$\frac{1}{{(2n + 3)(2n + 5)}} = \frac{1}{{4{n^2} + 16n + 15}} < \frac{1}{{{n^2}}}$$

because $4{n^2} + 16n + 15 > {n^2}$ for any $n \ge 1$ .

It means that if the series $\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^2}}}}$ converges, then the series $\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{(2n + 3)(2n + 5)}}}$ converges too.

To prove that $\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^2}}}}$ converges, we can use the Cauchy integral test: in short, the series $\sum\limits_{n = 1}^{ + \infty } {f(n)}$ converges or diverges simultaneously with the integral $\int\limits_1^{ + \infty } {f(x)dx}$. Thus, for $\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^2}}}}$ we get

$$\int\limits_1^{ + \infty } {\frac{1}{{{x^2}}}dx} = - \left. {\frac{1}{x}} \right|_1^{ + \infty } = 1$$

So the series $\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{{n^2}}}}$ converges and then the series

$$\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{(2n + 3)(2n + 5)}}}$$

converges too by the comparison theorem.