How prove this $|P_{n}(x)|\le 1$ let $0<r<1$,and 
 $$\dfrac{1}{\sqrt{1-2tr+r^2}}=\sum_{n=0}^{\infty}P_{n}(t)r^n$$
show that
$$|P_{n}(t)|\le 1,-1\le t\le 1$$
My try:
I know this coefficients $P_{n}$ are called Legendre polynomials,
$$P_{n}(t)=\sum_{k=0}^{[n/2]}\dfrac{n!}{2^{2k}(k!)^2(n-2k)!}t^{n-2k}(t^2-1)^k$$maybe prove $|P_{n}(t)|\le 1$ have some methods? Thank you for post you solution,Thank you

For $|t|\le 1$ we can put $t=\cos\theta$ so that we can write $$1-2tr+r^2=1-2r\cos\theta+r^2=\left(1-r\operatorname{e}^{+i\theta}\right)\left(1-r\operatorname{e}^{-i\theta}\right)$$
and using the expansion $(1+y)^m=\sum_{n=0}^{\infty}\binom{m}{n}x^n$ we have
$$
\begin{align}
\frac{1}{\sqrt{1-2r\cos\theta+r^2}}
&=\left(1-r\operatorname{e}^{+i\theta}\right)^{-1/2}\left(1-r\operatorname{e}^{-i\theta}\right)^{-1/2}\\
&=\sum_{m=0}^\infty a_mr^m\operatorname{e}^{+im\theta}\sum_{m=0}^\infty a_mr^m\operatorname{e}^{-im\theta}
\end{align}
$$
where $a_0=1$ and $a_m=\tfrac{1\cdot3\cdots(2m-1)}{2\cdot 4\cdots(2m)}$ for $m=1,2,\ldots$, so that
$$
\begin{align}
P_n(\cos\theta)&=\sum_{m=0}^n a_m\operatorname{e}^{+im\theta}a_{n-m}\operatorname{e}^{-i(n-m)\theta}\\
&=\sum_{m=0}^n a_m a_{n-m}\operatorname{e}^{-i(n-2m)\theta}\\
&=\sum_{m=0}^n a_m a_{n-m} \cos(n-2m)\theta\\
&=2a_0a_n\cos n\theta+2a_1a_{n-1}\cos (n-2)\theta+\cdots+
\begin{cases}2 a_{(n-1)/2}a_{(n+1)/2}\cos \theta& \text{for $n$ odd }\\
a^2_n/2 & \text{for $n$ even}
\end{cases}\\
&=\tfrac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots(2n)}\left(2\cos n\theta+\tfrac{1\cdot (2n)}{2\cdot (2n-1)}2\cos(n-2)\theta+\tfrac{1\cdot 3\cdot (2n)\cdot (2n-2)}{2\cdot 4\cdot(2n-1)\cdot(2n-3)}2\cos(n-4)\theta+\cdots\right)
\end{align}
$$
Consequently $P_n(\cos\theta)$ is a trigonometric cosine polynomial with non-negative coefficients.
If $\theta$ is a real angle
$$
|P_n(\cos\theta)|\le \tfrac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots(2n)}\left(2+\tfrac{1\cdot (2n)}{2\cdot (2n-1)}2+\tfrac{1\cdot 3\cdot (2n)\cdot (2n-2)}{2\cdot 4\cdot(2n-1)\cdot(2n-3)}2+\cdots\right)=P_n(1)
$$
so that $|P_n(\cos\theta)|\le 1$.
