Consider the series:
$$S(x)=\sum_{n=0}^{\infty}\frac{(2n+1)P_n(x)}{2}$$
Show that:
$$\int_{-1}^1 f(x)S(x)dx= f(1)$$
where $f(x)$ is any function of the interval $[−1, 1]$ on the real numbers which can be expressed as a series of Legendre polynomials.
Suppose $f$ can be expressed as a series of Legendre polynomials: $$ f(x)=\sum_{n=0}^{\infty}a_nP_n(x). \tag{1} $$ Then \begin{align} \int_{-1}^{1}f(x)S(x)\,dx&=\sum_{n=0}^{\infty}a_n\int_{-1}^{1}P_n(x)S(x)\,dx \\ &=\sum_{n=0}^{\infty}a_n\sum_{m=0}^{\infty}\frac{2m+1}{2}\int_{-1}^{1}P_n(x)P_m(x)\,dx \\ &=\sum_{n=0}^{\infty}a_n, \tag{2} \end{align} where the last equality follows from the orthogonality and normalization of the Legendre polynomials, $$ \int_{-1}^{1}P_n(x)P_m(x)\,dx=\frac{2}{2m+1}\,\delta_{nm}. \tag{3} $$ Since $P_n(1)=1$, it follows from $(1)$ that $f(1)=\sum_{n=0}^{\infty}a_n$. Comparing this result with $(2)$, we conclude that $$ \int_{-1}^{1}f(x)S(x)\,dx=f(1). \tag{4} $$