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I don't know how to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$,appreciate any help!

Is there any general rule for solving such problems?

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Call it $f$. Divide by $x$. Integrate term-by-term. Find closed form for result by thinking about $e^x$. Differentiate. Multiply by $x$. – Gerry Myerson Jun 20 '13 at 12:46
$$xe^{x^2}(2x^2+1)=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{n!}$$ – Ethan Jun 20 '13 at 12:56
There are general rules for expressing $$\sum \frac{p(n)}{n!}z^n$$ where $p$ is a polynomial, so you can take those answers and then do various substitutions to get this sort of answer. – Thomas Andrews Jun 20 '13 at 12:59
up vote 6 down vote accepted

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ $$xe^{x^2}=\sum_{n=0}^\infty\frac{x^{2n+1}}{n!}$$ $$\frac{d}{dx}xe^{x^2}=\sum_{n=0}^\infty\frac{(2n+1)x^{2n}}{n!}$$ $$xe^{x^2}(2x^2+1)=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{n!}$$

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I don't think there's a general rule but knowing some stuff can help:

$$f(x)=\sum_{n=0}^\infty\frac{2n+1}{n!}x^{2n}\implies\int f(x)\,dx=\sum_{n=0}^\infty\frac{x^{2n+1}}{n!}=x\sum_{n=0}^\infty\frac{(x^2)^n}{n!}=xe^{x^2}\ldots$$

Note that your series is $\,xf(x)\,\ldots$ , and also check where can you perform the above integration memberwise on that sum

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An alternate approach, without taking derivatives or anti-derivatives. Let $$g(z)=\sum_{n=0}^\infty \frac{2n+1}{n!}z^n$$ Then the function for the series you asked about is $xg(x^2)$.

But $$\sum \frac{2n+1}{n!}z^n = \sum \frac{2}{(n-1)!}z^n + \sum \frac{1}{n!}z^n = 2ze^z + e^z = (2z+1)e^z$$

So $xg(x^2) = x(2x^2+1)e^{x^2}$.

For a general polynomial $p(n)$ we can write $p$ as a linear combination of falling factorials, $(n)_k=n(n-1)\dots(n-k+1)$. So $p(n)=\sum_k a_k(n)_k$ for some finite sequence of values $a_k$ and then:

$$\sum_n \frac{p(n)}{n!} z^n = e^z \sum_k a_kz^k$$

For example, if $p(n)=n^2 = n(n-1) + n = (n)_2+(n)_1$, then $$\sum_n \frac{n^2}{n!}z^n = e^z(z^2+z)$$

Or $n^3 = (n)_3 + 3(n)_2 + (n)_1$ so: $$\sum_n \frac{n^3}{n!}z^n = e^z(z^3+3z^2+z)$$

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$$\text{As } e^x=\sum_{0\le r<\infty}\frac{x^r}{r!} $$ $$\text{and }\frac{(2n+1)}{n!}x^{2n+1}=2x^3 \cdot \frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \frac{(x^2)^n}{n!}$$

$$\text{So,} \sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$$ $$=2x^3 \cdot \sum_{n=0}^{\infty}\frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}$$

$$=2x^3 \cdot \sum_{m=0}^{\infty}\frac{(x^2)^m}{m!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}\text{ putting } n-1=m \text{ and using} \frac1{(-1)!}=0 $$

$$=2x^3 \cdot e^{x^2}+x\cdot e^{x^2}$$

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