Odd and Even Summation

$The\quad given\quad summation\quad is\quad equivalent\quad to:\\ \sum _{ n=1 }^{ \infty }{ \frac { ({ -1) }^{ n+1 }\times \left( 2n+1 \right) }{ 2n\times (2n+2) } } \\ =\frac { 1 }{ 4 } \sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n }\times (2n+1) }{ n(n+1) } } \\ =\frac { 1 }{ 4 } \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n }\frac { n+n+1 }{ n(n+1) } } \\ =\frac { 1 }{ 4 } \sum _{ n=1 }^{ \infty }{ { (-1) }^{ n }\left\{ \frac { 1 }{ n+1 } +\frac { 1 }{ n } \right\} } =\frac { 1 }{ 4 } \left\{ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } -\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } ....... \right\} \quad \left[ Telescoping\quad series \right] \\ =\frac { 1 }{ 4 } \left( 1 \right) =\boxed { 0.25 }$

Let the sum be $\mathfrak{C}$ . Multiply all numerators by $2$ .

$\mathfrak C=\dfrac{1}{2}\left(\dfrac6{2\cdot4}-\dfrac{10}{4\cdot6}+\dfrac{14}{6\cdot8}-\dfrac{18}{8\cdot10}+\cdots\right)$

We see numerators are sum of respective denominators so write them as $2+4;4+6;6+8;\cdots$ respectively and break into fractions so that $\mathfrak C$ is:

$\mathfrak C=\dfrac{1}{2}\left(\dfrac{2+4}{2\cdot4}-\dfrac{4+6}{4\cdot6}+\dfrac{6+8}{6\cdot8}-\dfrac{8+10}{8\cdot10}+\cdots\right)$

$\large 1/2\left(\dfrac{1}{2}+\color{#D61F06}{\dfrac{1}{4}-\dfrac{1}{4}}\color{#3D99F6}{-\dfrac{1}{6}+\dfrac{1}{6}}\color{teal}{+\dfrac{1}{8}}\cdots\right)$

$\Large =\dfrac12\times \left(\dfrac12\right)=\huge\boxed{0.25}$