Telescoping Problem

I have divided the question into two parts. One showing the simplified version of the summation, other showing the telescoping significance.

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Part I

$\large \displaystyle\sum _{ k=1 }^{ 2014 }{ \dfrac { k }{ \left( k+1 \right) ! } } =\dfrac { 1 }{ 2! } +\dfrac { 2 }{ 3! } +\dfrac { 3 }{ 4! } +...+\dfrac { 2014 }{ 2015! }$

Part 2

$\huge \left( \frac { 1 }{ 2! } +\frac { 2 }{ 3! } +\frac { 3 }{ 4! } +...+\frac { 2014 }{ 2015! } \right)+\frac { 1 }{ 2015! } =\frac { 1 }{ 2! } +\frac { 2 }{ 3! } +\frac { 3 }{ 4! } +..+\frac { 2014+1 }{ 2015! }$

$\huge =\frac { 1 }{ 2! } +\frac { 2 }{ 3! } +\frac { 3 }{ 4! } +...+\frac { 2013 }{ 2014! } +\frac { 1 }{ 2014! } =...=\frac { 1 +1}{ 2! } =1$

Note : The final answer follows from the fact that for each $k$ , we get an expression like $\dfrac{k}{(k+1)!}+ \dfrac {1}{(k+1)!} = \dfrac {1}{k!}$ which triggers a chain reaction and forms a telescoping series giving $1$ as the answer.