Sequences and S...

The sum of the series $$1+\dfrac{\log_e x}{1!}+\dfrac{(\log_e x)^2}{2!}+........$$ is 

If x<1, then $$\displaystyle \frac { 1 }{ 1+x } +\frac { 2x }{ 1+{ x }^{ 2 } } +\frac { { 4x }^{ 3 } }{ 1+{ x }^{ 4 } } +.........\infty =$$

Find the sum of 
$$1+\frac { 1 }{ 4 } +\frac { 1.3 }{ 4.8 } +\frac { 1.3.5 }{ 4.8.12 } +...\infty $$.

The sum of the series 1+2(1 +1/n)+ 3$${ \left( 1+1/n \right)  }^{ 2 }+....\infty \quad is\quad given\quad by$$

The $${ 7 }^{ th }$$ term of the series$$1+\dfrac { 1 }{ \left( 1+3 \right)  } { \left( 1+2 \right)  }^{ 2 }\dfrac { 1 }{ \left( 1+3+5 \right)  } { \left( 1+2+3 \right)  }^{ 2 }+.....,$$ is equal to ________.

The sum of the first n terms of the series $${ 1 }^{ 2 }+2.{ 2 }^{ 2 }+{ 3 }^{ 2 }+2.{ 4 }^{ 2 }+{ 5 }^{ 2 }+2.{ 6 }^{ 2 }+...$$ is $$\dfrac { n{ \left( n+1 \right)  }^{ 2 } }{ 2 } $$ when n is even. When n is odd the sum is

The sum of the series $$ 1 + \frac { 1 + 2 } { 2 } + \frac { 1 + 2 + 3 } { 3 } + \ldots $$ to n terms is

If $$\sum _{ k=2 }^{ n }{ cos{  }^{ -1 }(\frac { 1+\sqrt { (k-1)(k+2)(k+1)k }  }{ k(k+1) }  } )=\frac { 120\pi  }{ \lambda  }$$,then

Use geometric series to express $$0.555...=0.\overline 5$$ as a rational number.

If $$L=\sum _{ r=7 }^{ 2400 }{ \log _{ 7 }{ \left( \frac { r+1 }{ r }  \right)  }  } ,M=\prod _{ r=2 }^{ 1023 }{ \log _{ r }{ (r+1) }  } $$ and $$N=\sum _{ r=2 }^{ 2011 }{ \left( \frac { 1 }{ \log _{ r }{ p }  }  \right)  } $$ where p=$$(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...........\cdot 2011)$$, then