[SOLVED] Binomial Theorem(Sum of series)?

• Apr 23rd 2009, 05:02 AM
fardeen_gen
[SOLVED] Binomial Theorem(Sum of series)?
Find the sum of the series:
1 + {(√2 - 1)/(2√2)} - {(2√2 - 3)/(12)} + {(5√2 - 7)/(48√2)} - {(12√2 - 17)/(480)} + ... upto ∞ terms.
• Apr 23rd 2009, 06:12 AM
Series
Hello fardeen_gen
Quote:

Originally Posted by fardeen_gen
Find the sum of the series:
1 + {(√2 - 1)/(2√2)} - {(2√2 - 3)/(12)} + {(5√2 - 7)/(48√2)} - {(12√2 - 17)/(480)} + ... upto ∞ terms.

Note that:

• $\displaystyle (\sqrt2-1)^2 = (3-2\sqrt2)$

• $\displaystyle (\sqrt2-1)^3 = (5\sqrt2-7)$

• $\displaystyle (\sqrt2-1)^4 = (17 -12\sqrt2)$

So, if $\displaystyle x = (\sqrt2-1)$ and we make all the signs positive, the numerators are $\displaystyle 1, x, x^2, x^3, x^4, ...$

The denominators are: $\displaystyle 1, 1 \cdot2\sqrt2, 1 \cdot2\sqrt2\cdot 3\sqrt2,1 \cdot2\sqrt2\cdot3\sqrt2\cdot4\sqrt2,1 \cdot2\sqrt2\cdot3\sqrt2\cdot4\sqrt2\cdot5\sqrt2, ...$

i.e. $\displaystyle 1, 2!\sqrt2, 3!(\sqrt2)^2,4!(\sqrt2)^3,5!(\sqrt2)^4, ...$

So if we let $\displaystyle y =\frac{x}{\sqrt2}= \frac{\sqrt2-1}{\sqrt2}$ the series is

$\displaystyle S=1 + \frac{y}{2!}+\frac{y^2}{3!}+\frac{y^3}{4!}+\frac{y ^4}{5!}+ ...$

$\displaystyle \Rightarrow 1+ Sy = 1 +y+ \frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+\frac {y^5}{5!}+ ...$

= $\displaystyle e^y$

$\displaystyle \Rightarrow S = \frac{e^y-1}{y}$, where $\displaystyle y = \frac{\sqrt2-1}{\sqrt2}$