Binomial Series — Question 10

Question 10 – Solution

Step 1: Substitute into the parent series.

$$(1+u)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}u^n.$$ With $u=3x$, $$\boxed{(1+3x)^{-1/2}=\sum_{n=0}^{\infty}\binom{-1/2}{n}3^n x^n=1-\frac32x+\frac{27}{8}x^2-\frac{135}{16}x^3+\cdots}.$$

Step 2: Find the radius.

The condition $|u|<1$ becomes $|3x|<1$, so $R=1/3$.

Step 3: Test endpoints.

At $x=1/3$, $u=1$, and the terms alternate with magnitude asymptotic to $1/\sqrt{\pi n}$; the series converges conditionally. At $x=-1/3$, $u=-1$ cancels the coefficient signs, leaving positive terms comparable with $1/\sqrt n$; it diverges.

The endpoint magnitudes $b_n=4^{-n}\binom{2n}{n}$ decrease, since $b_{n+1}/b_n=(2n+1)/(2n+2)<1$, and tend to zero. This verifies the alternating-test hypotheses.

Conclusion.

The interval is $\boxed{(-1/3,1/3]}$.