I am trying out to solve a differential equation using a power series solution.
I figured out the recurrence or recursive relation of the constants and I figured out the pattern somehow.
By substituting n values to the recurrence relation, I got a pattern like this.
Recurrence relation:
\begin{align}
a_{n}&=\frac{-(-5+2n)a_{n-1}}{(2n+(4(n^{2})))}
\\ \\
a_{n}&=\frac{(-(-1))(-(-1))(-3)(-3)(-5)(-5)...(-(-5+2n)) a_{0}}{(1)(2)(3)(4)(5)(6)(7)...(2n+4n^{2})}\end{align}
I kind off got how it goes like it has a pattern of odd numbers at the power of 2 and in the denominator is just n! so I got a power series out of it.
$$ \sum_{n=1}^{\infty} \frac{((-1)^{n+1})(2n+1)^{2})(a_{0})(x^n)}{(n!)(2n+4n^{2})}$$
To summarize it all, the expression $(2n+1)^{2}$ was to replace the odd numbers at the power of $2$ and the $n!$ to replace the factorial of numbers. And I just left the recurrence or recursive relation. The summation goes to infinity and has an index of $1$ since it had the index of $1$ even before I got the n-values.
Is the power series representation correct?
Assuming the recurrence is $$ a_{n}=\frac{-(2n-5)a_{n-1}}{2n(2n+1)}$$ multiply by $$n!(-2)^n(2n+1)(2n-1)(2n-3)$$ (using the recursive relations $-2=(-2)^{n}/(-2)^{n-1}$, $n=n!/(n-1)!$ etc.) to get \begin{align} b_n&=(-2)^nn!(2n+1)(2n-1)(2n-3)a_n\\ &=(-2)^{n-1}(n-1)!(2n-1)(2n-3)(2n-5)\\ &=b_{n-1} \end{align} so that the $b$ sequence is constant with $b_0=3a_0$ and thus $$ a_n=\frac{3(-2)^{-n}a_0}{n!(2n+1)(2n-1)(2n-3)}. $$
Or in another way, $a_N/a_0$ is the product of all those fractions from $n=1$ to $N$, or the fraction of the products of the single factors, $$ \frac{a_N}{a_0}=\frac{(-1)^N\prod_{n=1}^N(2n-5)}{2^NN!\prod_{n=1}^N(2n+1)} $$ Now it takes some observation to remark that most of the odd numbers in the last factors cancel, as $2n+1=2(n+3)-5$. Thus there remains $$ \frac{a_N}{a_0}=\frac{(-1)^N\prod_{n=1}^3(2n-5)}{2^NN!\prod_{n=N-2}^N(2n+1)} =\frac{(-1)^N(-3)(-1)(1)}{2^NN!(2N-3)(2N-1)(2N+1)}. $$