Sum of N terms of a series

Question

Perhaps this is a silly question but how can I simplify this to find the sum of $N$ terms?

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

I know how to find the sum of cubes and squares etc but how with the denominator like that? Any help would be appreciated thanks.

The questions in the book have all involved finding things like:

$$\sum_{n=1}^N {(3n^3+n^2+n)}$$

This is the only one with the cubes and squares in the denominator

Answer 1

$\sum\limits_{n=1}^{N}\dfrac{2n-1}{n(n+1)(n+2)}=\sum\limits_{n=1}^{\infty}\dfrac{2n-1}{n(n+1)(n+2)}-\sum\limits_{n=N+1}^{\infty}\dfrac{2n-1}{n(n+1)(n+2)}=\dfrac{3}{4}-\dfrac{4N+3}{2N^2+6N+4}$

Answer 2

Checking if there is a pair of constants $(A,B)$ such that $$\frac{2n-1}{n(n+1)(n+2)}=A\cdot\color{red}{\frac{1}{(n+1)(n+2)}}+B\cdot\color{blue}{\frac{1}{n(n+1)}}$$ gives us $(A,B)=(5/2,-1/2)$.

Hence, since we can have

$$\frac{2n-1}{n(n+1)(n+2)}=\frac{5}{2}\color{red}{\left(\frac{1}{n+1}-\frac{1}{n+2}\right)}-\frac 12\color{blue}{\left(\frac{1}{n}-\frac{1}{n+1}\right)},$$ we have $$\sum_{n=1}^{N}\frac{2n-1}{n(n+1)(n+2)}=\frac{5}{2}\sum_{n=1}^{N}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)-\frac 12\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ $$=\frac{5}{2}\left\{\left(\frac 12-\frac 13\right)+\left(\frac 13-\frac 14\right)+\cdots+\left(\frac 1N-\frac{1}{N+1}\right)+\left(\frac{1}{N+1}-\frac{1}{N+2}\right)\right\}$$$$-\frac 12\left\{\left(\frac 11-\frac 12\right)+\left(\frac 12-\frac 13\right)+\cdots+\left(\frac{1}{N-1}-\frac{1}{N}\right)+\left(\frac 1N-\frac{1}{N+1}\right)\right\}$$ $$=\frac 52\left(\frac 12-\frac{1}{N+2}\right)-\frac 12\left(\frac 11-\frac{1}{N+1}\right)=\frac{N(3N+1)}{4(N+1)(N+2)}.$$

Answer 3

HINT:$$\dfrac{1}{(n+1)(n+2)}=\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}$$ and $$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2n(n+1)}-\dfrac{1}{2(n+1)(n+2)}$$

Answer 4

Split this into two parts, each with a simple numerator, which is more convenient for converting to partial fractions.

The second last line is probably more useful if taking limits as $N\to\infty$ is required.

Check:

Put $N=1$.

LHS = $\dfrac 1{1\cdot 2\cdot 3}=\dfrac 16$

RHS =$\dfrac {1\cdot 4}{4\cdot 2\cdot 3}=\dfrac 16$

Looks OK.