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Let us call the roots c_j for 1\leq j\leq m. If we take the product of two exponentials with the same base, we simply add the exponents: For the function f (x) = xn, n should not equal 0, for reasons which will become clear.
So, the series converges when $|x| Webfirst remark that the polynomial x^m+k has simple roots. Webwhat is the derivative of xn? Therefore x^n/(x^m+k) rewrites as a sum of simple elements. Xaxb = xa + b. Product of exponentials with same base. Webusing the ratio test, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)x^{n+1}}{nx^n} = \dfrac{(n+1)x}{n}$.
Xaxb = xa + b. Product of exponentials with same base. Webusing the ratio test, $\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)x^{n+1}}{nx^n} = \dfrac{(n+1)x}{n}$. $$f(x) = \sum_{n=1}^\infty n.