Not Double Factorial

Algebra Level 3

$\frac { 1 }{ 1 \times 3 } +\frac { 2 }{ 1 \times 3 \times 5 } +\frac { 3 }{ 1 \times 3 \times 5 \times 7 } + \frac { 4 }{ 1 \times 3 \times 5 \times 7 \times 9 } +\ldots$

Find the sum of the series above.

The answer is 0.5.

1 solution

Mudit Bansal
Jan 19, 2015

I can rewrite the sum as: $\frac { 1 }{ 2 } \sum _{ r=1 }^{ \infty }{ \frac { \left( 2r+1 \right) -1 }{ 1.3.5....\left( 2r+1 \right) } } \\ =\quad \frac { 1 }{ 2 } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ 1.3.5...\left( 2r-1 \right) } -\frac { 1 }{ 3.5...\left( 2r+1 \right) } }$ now on applying the summation we get: $=\quad \frac { 1 }{ 2 } \left[ (\frac { 1 }{ 1 } -\frac { 1 }{ 3 }) +(\frac { 1 }{ 3 } -\frac { 1 }{15}) + \ldots \right]$ as we can see the term cancels out and the final answer is $\frac { 1 }{ 2 } *\frac { 1 }{ 1 }$ $=\quad 0.5$

As the series tends to Infinity, the denominator will become infinite and the last term will be zero. So, the result is 0.5

AMAN KUMAR - 6 years, 4 months ago

Nice solution!

But it appears there are some typos.

Here is the corrected solution:

I can rewrite the sum as: $\frac { 1 }{ 2 } \sum _{ r=1 }^{ \infty }{ \frac { \left( 2r+1 \right) -1 }{ 1\cdot3\cdot5...\left( 2r+1 \right) } } \\ =\quad \frac { 1 }{ 2 } \sum _{ r=1 }^{ \infty } \left [ { \frac { 1 }{ 1\cdot3\cdot5...\left( 2r-1 \right) } -\frac { 1 }{ 1\cdot3\cdot5...\left( 2r+1 \right) } } \right]$

now on applying the summation we get: $=\quad \frac { 1 }{ 2 } \left[ \frac { 1 }{ 1 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -\frac { 1 }{ 15 } +\frac { 1 }{ 15 } -......upto\quad infinity \right]$

as we can see the series telecopes and the final answer is $\frac { 1 }{ 2 } *\frac { 1 }{ 1 }$ $=\quad 0.5$

Michael Fischer - 6 years, 4 months ago

Thanks for reminding about the typos I've corrected them.

mudit bansal - 6 years, 4 months ago

Not Double Factorial

The answer is 0.5.

1 solution

0 pending reports