Answer
$\sum_{k=0}^{\infty} \dfrac{(-1)^k \ (x^{2k+1})}{(k+1)^2}$
Work Step by Step
We are given the power series $x-\dfrac{x^3}{4}+\dfrac{x^5}{9}-\dfrac{x^7}{16}+...$
or, $x-\dfrac{x^3}{4}+\dfrac{x^5}{9}-\dfrac{x^7}{16}+...=\dfrac{(-1)^0 \cdot x^{(2)(0)+1}}{(0+1)^2}+\dfrac{(-1)^1 \cdot x^{(2)(1)+1}}{(1+1)^2}......$
The given series can be represented in the summation form as:
$x-\dfrac{x^3}{4}+\dfrac{x^5}{9}-\dfrac{x^7}{16}+...=\sum_{k=0}^{\infty} \dfrac{(-1)^k \ (x^{2k+1})}{(k+1)^2}$
