If $\sum_{n=1}^\infty {6 \over n(n+3)}$ converges find its sum

Question

I'm guessing this is a duplicate but I don't understand the answers to this question (which is very similar to mine). We have only gone over the nth term test and the ratio test in my class so far, so I am assuming I should be able to solve this problem with one of those.

So I tried the nth term test:

$\lim_{n \to \infty} {6 \over n(n+3)} = 0$

Which means that test is inconclusive

Now the ratio test:

$\begin{align} \lim_{n \to \infty}\left|{6 \over (n+1)(n+4)} \cdot {n(n+3) \over 6}\right| \\ & = \lim_{n \to \infty}\left|{n^2 + 3 \over n^2 + 5n +5}\right| \\ & = 1 \end{align}$

Which is also inconclusive? This is where I am stuck, and I also don't know how to sum that series if it is convergent since when I write the first few terms out I can't seem to find a pattern like I usually do in an infinite series.

$6[\frac 1 4 + \frac 1 {10} + \frac 1 {18} + \frac 1 {28} + ... ]$

a = ? r = ?

I probably messed the tests up somehow?

Answer 1

You have directly $$ \frac{6}{n\left(n+3\right)} \underset{+\infty)}{\sim}\frac{6}{n^2} $$ The series $\displaystyle \sum_{n \geq 1}\frac{1}{n^2}$ converges, so as your series. To find the sum, can you find $\alpha$ and $\beta$ such as $$ \frac{1}{n\left(n+3\right)}=\frac{\alpha}{n}+\frac{\beta}{n+3} \ \ \ ? $$ And then calculate $$ S_n=\sum_{k=1}^{n}\frac{6}{k\left(k+3\right)} $$

Careful, your sum does not exist for $n=0$.

Answer 2

Hint. Observe that by partial fraction decomposition one gets $$ \frac6{n(n+3)}=\frac2{n}-\frac2{n+3} $$ giving by telescoping terms

$$ \sum_{n=1}^N\frac6{n(n+3)}=\frac{11}3-\frac2{N+1}-\frac2{N+2}-\frac2{N+3} $$

this yields both convergence of the given series and a closed form for its sum.

Answer 3

Hint:

To find the sum, rewrite the general term as follows: $$\frac 6{n(n+3)}=\frac 2n-\frac2{n+3}$$ You obtain a telescoping sum for each group of 4 terms, so you can find a closed form for the partial sums.

Answer 4

As shown by others, your series converges. To find its sum we can always use our favourite alternative method of converting the sum into a double integral.

Noting that $$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^{n + 2} \, dy = \frac{1}{n + 3},$$ the sum can be written as \begin{align*} \sum_{n = 1}^\infty \frac{6}{n (n + 3)} &= 6 \sum_{n = 1}^\infty \int_0^1 \int_0^1 x^{n - 1} y^{n + 2} \, dx dy\\ &= 6 \int_0^1 \int_0^1 \sum_{n = 1}^\infty x^{n - 1} y^{n + 2} \, dx dy \tag1\\ &= 6 \int_0^1 \int_0^1 \frac{y^2}{x} \sum_{n = 1}^\infty (xy)^n \, dx dy\\ &= 6 \int_0^1 \int_0^1 \frac{y^2}{x} \cdot \frac{xy}{1 - xy} \, dx dy \tag2\\ &= 6 \int_0^1 \int_0^1 \frac{y^3}{1 - xy} \, dx dy\\ &= -6 \int_0^1 y^2 \ln (1 - xy) \Big{|}_0^1 \, dy\\ &= -6 \int_0^1 y^2 \ln (1 - y) \, dy\\ &= 2 \int_0^1 \frac{1 - y^3}{1 - y} \, dy \tag3\\ &= 2 \int_0^1 (1 + y + y^2) \, dy \tag4\\ &= 2 \left [y + \frac{y^2}{2} + \frac{y^3}{3} \right ]_0^1\\ &= 2 \left (1 + \frac{1}{2} + \frac{1}{3} \right )\\ &= \frac{11}{3}. \end{align*}

Explanations

1. Interchanging the sum with the double integration.

2. Summing the series which is geometric.

3. Integrating by parts.

4. Using the factorisation $1 - y^3 = (1 - y)(1 + y + y^2)$.