Primary consecutive addition (Part 7)

The summation can be expressed as $\sum_{n=1}^{50}\frac{n}{(n+1)(n+2)} = \sum_{n=1}^{50}\frac{n}{n+1}-\frac{n}{n+2}$

Expanding this we notice that the harmonic series emerges: $\Large{\frac{1}{2} \underbrace{\color{#D61F06}{- \frac{1}{3} + \frac{2}{3}}}_{\frac{1}{3}} \underbrace{\color{#3D99F6}{ - \frac{2}{4} + \frac{3}{4}}}_{\frac{1}{4}} \cdots \underbrace{\color{#20A900}{ - \frac{48}{50} + \frac{49}{50}}}_{\frac{1}{50}} \underbrace{\color{#EC7300}{ - \frac{49}{51} + \frac{50}{51}}}_{\frac{1}{51}} - \frac{50}{52}}$

$\Large{=\color{#D61F06}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} \cdots +\frac{1}{50}+\frac{1}{51}} - \frac{50}{52}}$

$\therefore$ The summation is equal to the sum of the harmonic series up to the 51st term minus 1 minus $\frac{50}{52}$ .

Note that an approximation for the sum of the harmonic series up to the nth term is $\large{\ln(n) + \gamma + \frac{1}{2n} - \frac{1}{12n^{2}}}$ where $\gamma$ is the Euler-Mascheroni constant and is approximated by $\large{\frac{\pi}{2e}}$ .

So our final step is:

$\large{\ln(51) + \gamma + \frac{1}{102} - \frac{1}{12\times 51^{2}} - 1 - \frac{50}{52} \approx 2.557}$

$The~series~is~~\sum_1^{50}\dfrac n {(n+1)*(n+2)}\\ =\sum_1^{50}\dfrac 2 {n+2} - \sum_1^{50}\dfrac 1 {n+1}\\ =2*\sum_1^{50}\dfrac 1 {n+2} - \dfrac 1 2 -\sum_2^{50}\dfrac 1 {n+1}\\ =2*\sum_1^{49}\dfrac 1 {n+2}+\dfrac 2{52}- \dfrac 1 2 -\sum_1^{49}\dfrac 1{n+2}\\ =\sum_1^{49}\dfrac 1 {n+2}+\dfrac 2{52} - \dfrac 1 2\\ =\dfrac 1{26} - \dfrac 1 2 + \sum_3^{51}\dfrac 1 n=\Large ~~~~\color{#D61F06}{ 2.55727}$