# Solution of Equation $1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2}$

I have an equation whose left side is a infinite series . I can solve the equation if I am able to find a close form of the series . The equation is as follows :

$$1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2}$$

I have tried to find the sum of the series by trying to apply binomial theorm of negative exponent . But I cant solve this problem . Can you help me to find the sum of series of the left hand side of the above equation ?

• What are the denominators of the coefficients? – awwalker Sep 30 '13 at 6:07
• I dont understand your question . – Way to infinity Sep 30 '13 at 6:11
• It appears that there was a typo in your formula. The general case is now clear to me. – awwalker Sep 30 '13 at 6:12

Compare the series with $1+na+\frac{n(n-1)a^2}{2!}\cdots$ we get

$na=x$ or $$n^2a^2=x^2\dots(1)$$

also $$\frac{n(n-1)a^2}{2!}=\frac{1.3x^2}{2!}\cdots(2)$$

Divide $(1)$ and $(2)$ and solve for $n$, substitute in first and solve for $a$

you will get the sum to be $(1-2x)^{\frac{-1}2}$

Hint: Find the power series expansion of $(1-2x)^{-1/2}$.

• I reached the same answer using Mathematica, after finding the general form for the $n$th coefficient. I'd like to know: did you find your answer in a similar way? – awwalker Sep 30 '13 at 6:21
• No, can't afford Mathematica, it was by hand. – André Nicolas Sep 30 '13 at 6:22