# Suppose $\sum_{n=1}^{\infty} b_n x^n= \frac{x^3}{(x^4-1)^2}$. What could be an expression of $b_n$?

Suppose $$\sum_{n=1}^{\infty} b_n x^n = \frac{x^3}{(x^4-1)^2}.$$

What could be an expression of $b_n$?

Some of the possible answers read $2^{3n}nx^{3n-1}, nx^{3n-1}, nx^{4n-1}$

There are a few other answers. I was just not sure how to start this problem?

• Have you heard of Taylor series? That could be one place to start. – Matt Apr 27 '17 at 18:49
• the expression depends on the value of $x$ because the radius of convergence of a geometric series is $1$. – Masacroso Apr 27 '17 at 18:51
• Hint $\sum_{n=1}^{\infty} ny^{n} =\frac{y}{(1-y)^2}$ ... $y \rightarrow x^{\color{red}{?}}$ ... – Donald Splutterwit Apr 27 '17 at 18:53

HINT:

$$\int \frac{x^3}{(x^4-1)^2}\,dx=\frac{1}{4(1-x^4)}+C$$

Expand $\frac{1}{1-z}$ in a geometric series and set $z=x^4$?

• Does that mean I can say $$\int \sum_{n=1}^{\infty} b_n x^n = \frac{1}{4(1-x^4)} + C$$ could be written as a power function $$\sum z^n,$$ where z = x^2?  – Haim Apr 27 '17 at 19:59
• Note quite. Note $z=x^4$. – Mark Viola Apr 27 '17 at 20:08
• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola May 1 '17 at 3:48

Another variation is based upon the binomial series expansion. \begin{align*} (1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1 \end{align*}

We obtain \begin{align*} \frac{x^3}{(x^4-1)^2}&=x^3\sum_{n=0}^\infty\binom{-2}{n}(-x^4)^n\tag{1}\\ &=x^3\sum_{n=0}^\infty\binom{n+1}{1}x^{4n}\\ &=\sum_{n=0}^\infty (n+1)x^{4n+3}\tag{2}\\ &=x^3 + 2 x^7 + 3 x^{11} + 4 x^{15} + 5 x^{19} + 6 x^{23}+\cdots \end{align*}

In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

We conclude according to (2) the representation \begin{align*} \sum_{n=1}^{\infty} b_n x^n=\sum_{n=0}^\infty (n+1)x^{4n+3} \end{align*} implies for $n\geq 0$ \begin{align*} b_{4n+k}= \begin{cases} n+1&\qquad k=3\\ 0&\qquad k\neq 3 \end{cases} \end{align*}