关于含double factorial的无穷级数奇数项等价性的证明问询

Hey there, let's work through this step by step to connect the odd double factorial series to that tricky-looking expression you've proposed. First, let's fix a small notation issue upfront: you used $n$ as the index for both the outer and inner sums, which is a typo—let's switch the outer sum to use $k$ instead to avoid confusion.

Step 1: Rewrite the inner sums using integral representations #

First, let's tackle the two inner series in your expression. A handy trick for terms with denominators like $(4n+1)$ or $(4n+3)$ is to use the integral identity $\frac{1}{m} = \int_0^1 t^{m-1} dt$ (valid for positive $m$). This lets us convert the series into integrals of power series, which are easier to simplify.

For the first inner series:
$$
\sum_{n=0}^{\infty} \frac{2^{-2n}}{(4n+1)(2n)!} = \sum_{n=0}^{\infty} \frac{2^{-2n}}{(2n)!} \int_0^1 t^{4n} dt
$$
Since the series converges absolutely (we can verify this with the ratio test), we can swap the sum and integral:
$$
= \int_0^1 \sum_{n=0}^{\infty} \frac{(t^4 / 4)^n}{(2n)!} dt
$$
Recall the hyperbolic cosine series: $\cosh(z) = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}$. If we let $z = t^2/2$, then $z^{2n} = t^{4n}/4n$, so the inner sum simplifies to $\cosh\left(\frac{t^2}{2}\right)$. So this first inner series becomes:
$$
\int_0^1 \cosh\left(\frac{t^2}{2}\right) dt
$$

Now for the second inner series:
$$
\sum_{n=0}^{\infty} \frac{2^{-(2n+1)}}{(4n+3)(2n+1)!} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^{-2n}}{(2n+1)!} \int_0^1 t^{4n+2} dt
$$
Again, swap sum and integral:
$$
= \frac{1}{2} \int_0^1 t^2 \sum_{n=0}^{\infty} \frac{(t^4 / 4)^n}{(2n+1)!} dt
$$
Use the hyperbolic sine series: $\sinh(z) = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!}$. Rearranging gives $\sum_{n=0}^{\infty} \frac{z^{2n}}{(2n+1)!} = \frac{\sinh(z)}{z}$. Letting $z = t^2/2$ again, the inner sum becomes $\frac{2 \sinh\left(\frac{t^{2}{2}\right)}{t}2}$. Multiply by $t^2$ and the $1/2$ factor, and this inner series simplifies to:
$$
\int_0^1 \sinh\left(\frac{t^2}{2}\right) dt
$$

Step 2: Simplify the difference of the inner sums #

Now subtract the two simplified inner series:
$$
\int_0^1 \cosh\left(\frac{t^2}{2}\right) dt - \int_0^1 \sinh\left(\frac{t^2}{2}\right) dt = \int_0^1 \left( \cosh\left(\frac{t^2}{2}\right) - \sinh\left(\frac{t^2}{2}\right) \right) dt
$$
Remember that $\cosh(x) - \sinh(x) = e^{-x}$, so this reduces nicely to:
$$
\int_0^1 e^{-t2/2} dt
$$

Step 3: Evaluate the outer sum #

The outer sum in your expression is:
$$
\sum_{k=0}^{\infty} \frac{(1/2)^k}{k!}
$$
This is just the Taylor series for $e^x$ evaluated at $x = 1/2$, so it equals $e^{1/2} = \sqrt{e}$.

Putting it all together, your original complex expression simplifies to:
$$
\sqrt{e} \times \int_0^1 e^{-t2/2} dt
$$

Step 4: Connect to the odd double factorial series #

Now we just need to show the odd double factorial series equals this same expression. Let's start with a standard identity for the odd double factorial series. Consider the power series:
$$
\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!!}
$$
We can prove this equals $e^{x2/2} \int_0^x e^{-t2/2} dt$ by verifying they satisfy the same differential equation and initial condition:

• Take the derivative of the right-hand side: $\frac{d}{dx} \left( e^{x2/2} \int_0^x e^{-t2/2} dt \right) = x e^{x2/2} \int_0^x e^{-t2/2} dt + 1$
• Take the derivative of the left-hand side: $\sum_{n=0}^{\infty} \frac{(2n+1)x^{2n}}{(2n+1)!!} = 1 + x \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!!}$

Both derivatives follow the equation $y' = x y + 1$, and both equal 0 when $x=0$, so they must be identical. Now set $x=1$:
$$
\sum_{n=0}^{\infty} \frac{1}{(2n+1)!!} = e^{1/2} \int_0^1 e^{-t2/2} dt
$$

This is exactly the simplified form of your proposed expression! That's the equivalence you were looking for.

As a quick side note: the connection to Fresnel integrals you noticed makes sense because $\int e^{-t2/2} dt$ can be related to Fresnel integrals via variable substitutions, but that's a separate rabbit hole. For this equivalence proof, the integral and hyperbolic function tricks are all you need.

备注：内容来源于stack exchange，提问作者Chicago