Infinite summation question and i need to make an algorithm for finding the summation?

Question

$$\sum_{n=1}^{\infty} \frac {x^{2n -1}} {2n!}$$

for the algorithm i use == $d=\frac {a_{n}} {a_n - 1}$

And other hint that I have is the for $\sum_{n=1}^{\infty} \frac {x^{2n}} {n!}$ ; $d = \frac{x}{n} $

So my question is that what will be d = ? for $\sum_{n=1}^{\infty} \frac {x^{2n -1}} {2n!}$

Answer 1

I think you want a recursive relation for your algorithm or a similar thing, so perhaps $$d=\dfrac{a_{n}}{a_{n-1}}=\dfrac{x^2}{n}$$ works, or want to have $d=s_{k}-s_{k-1}=\dfrac{x^{2k-1}}{2k!}$ where $$s_k=\sum_{n=1}^{k} \frac {x^{2n -1}} {2n!}$$ is partial sum of the series. In such case the relation is $$s_{k}=\dfrac{x^{2k-1}}{2k!}+s_{k-1}$$ where $s_1=\dfrac{x}{2}$.

Answer 2

You can compute the sum with the exponential function: $$\sum_{n=1}^{\infty} \frac {x^{2n -1}} {2n!} =\frac{1}{2x}\sum_{n=1}^{\infty} \frac {x^{2n}} {n!} =\frac{1}{2x}\sum_{n=1}^{\infty} \frac {(x^2)^n} {n!} = \frac{e^{x^2}-1}{2x}$$ here used the well-known formula for $e^y = \sum_{n=0}^{\infty} \frac {y^n} {n!}$and therefore $$ \sum_{n=1}^{\infty} \frac {y^n} {n!} = e^y -1$$