It's Definitely Not Geometric

$\displaystyle S =1 + \frac {1}{2} + \frac {1}{6} + \frac {1}{12} + \frac {1}{20} + ... = 1 + \sum_{n=1}^\infty {\frac {1}{n(n+1)}}$

$\displaystyle \quad = 1 + \sum_{n=1}^\infty {\left( \frac {1}{n} - \frac {1}{n+1}\right)} = 1 + \sum_{n=1}^\infty {\frac {1}{n}} - \sum_{n=1}^\infty {\frac {1}{n+1}}$

$\displaystyle \quad = 1 + \sum_{n=1}^\infty {\frac {1}{n}} - \sum_{n=2}^\infty {\frac {1}{n}} = 1 + \sum_{n=1}^1 {\frac {1}{n}} = 1 + 1 = \boxed{2}$

I solved this the same way as Chew-Seong Cheong. Here it is for level 2 solvers with a few more details and without using the summation notation.

First of all identify the pattern for the denominators and write the series as

$1+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...$

Now use the method of partial fractions to write each fraction as the sum of two fractions. If you don't know how to do this you can still see why the following line is correct by putting each pair of fractions back over their common denominator

$1+(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+...$

Now notice how the last term in each bracket cancels the first term in the next bracket, so the whole series collapses like a folding telescope to give the answer

$1+1=\boxed{2}$

The question can also be written as

$1+\bigg (1-\frac { 1 }{ 2 } \bigg )+\bigg (\frac { 1 }{ 2 } -\frac { 1 }{ 3 } \bigg )+\bigg (\frac { 1 }{ 3 } -\frac { 1 }{ 4 } \bigg )+\bigg (\frac { 1 }{ 4 } -\frac { 1 }{ 5 } \bigg )\dots$

$\Rightarrow 1+1+\frac { 1 }{ \infty }$ but $\frac { 1 }{ \infty } =0$

$\therefore$ the answer is $2$ .

Notice that the sequence can be rewritten as follows.

$1+ \frac {1}{2}(1 + \frac {1}{3} + \frac {1}{6} + \frac {1}{10} + . . . )$

This remind you of anything? Triangular numbers?

$= 1 + \frac {1}{2} \left( \displaystyle \sum_{n=1}^{\infty} \frac {2}{n(n+1)} \right)$

$= 1 + \displaystyle \sum_{n=1}^{\infty} \frac {1}{n(n+1)}$

$= 1 + \displaystyle \sum_{n=1}^{\infty} \frac {1}{n} - \frac {1}{n+1}$

$= 1 + \left ( \frac {1}{1} - \frac {1}{2} + \frac {1}{2} - \frac {1}{3} + \frac {1}{3} - \frac {1}{4} + . . . - \frac {1}{\infty} \right)$

$= 1 + 1 - 0 = \boxed {2}$

1 - 1/2 = 1/2

1/2 - 1/3 = 1/6

1/3 - 1/4 = 1/12

1/4 - 1/5 = 1/20

1/n - 1/(n + 1 ) = 1/n(n + 1)

Then

1/2 +1/6 + 1/12 + 1/20 + .... + 1/n(n + 1) = 1 - 1/(n + 1) = n/(n + 1)

Lim {n/(n + 1)} = 1 when n tends to infinity

Then

1/2 +1/6 + 1/12 + 1/20 + .... = 1

So

1 + 1/2 +1/6 + 1/12 + 1/20 + .... = 2

We can write it like 1+1/2+(1/2-1/3)+(1/3-1/4)+(1/4-1/5).........so first 3 terms will remain and all other terms get cancelled. so the sum of the series will come out to be 2.

1 + (1 - 1/ 2) + (1/ 2 - 1/ 3) + (1/ 3 - 1/4) + (1/ 4 - 1/ 5) + .... = 2.

This is found in Cayley Series of ellipse perimeter. (1/ 0 - 1) is not expressed.