A nifty series involving $\cosh(x)$

Question

There are many series that can be found in the literature that are entertaining. Here is yet another.

What is the resulting value of the series $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \left( \cosh(ax) - \sum_{r=0}^{n} \frac{(ax)^{2r}}{(2r)!} \right)$$

Answer 1

Reverse the order of summation:

$$ \begin{align}\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \left( \cosh(ax) - \sum_{r=0}^{n} \frac{(ax)^{2r}}{(2r)!} \right) &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \sum_{r=n+1}^{\infty} \frac{(ax)^{2r}}{(2r)!}\\ &= \sum_{r=1}^{\infty} \frac{(ax)^{2r}}{(2r)!} \sum_{n=1}^r \frac{(-1)^{n+1} \, n^{2}}{2^{n}}\end{align}$$

The inner sum is simply a twice-differentiated geometric sum:

$$\sum_{n=1}^r \frac{(-1)^{n+1} \, n^{2}}{2^{n}} = \frac{1}{27} \left[2-\left(-\frac{1}{2}\right)^r (9 r^2+12 r+2)\right] $$

Now the outer sum may be evaluated using standard geometric series techniques.

Answer 2

Hint: First, simplify your life by writing $\dfrac{(-1)^{n+1}n^2}{2^n}=-n^2\bigg(-\dfrac12\bigg)^n$ and $n^kt^n=\bigg(t~\dfrac d{dt}\bigg)^kt^n$.

This, together with the $($other$)$ hint provided in the comments, and the reversing of the order of the two summations, yields the desired result.

Answer 3

Let $$S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \left( \cosh(ax) - \sum_{r=0}^{n} \frac{(ax)^{2r}}{(2r)!} \right)$$ By using \begin{align} \sum_{n=0}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} &= \frac{2}{27} \\ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} \, (n+r)^{2}}{2^{n}} &= \frac{2}{27} \, (1 + 6 r - 9 r^{2}) \end{align} then the series is evaluated as follows: \begin{align} S &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \left( \cosh(ax) - \sum_{r=0}^{n} \frac{(ax)^{2r}}{(2r)!} \right) \\ &= \frac{2}{27} \, \cosh(ax) - \sum_{n=0}^{\infty} \sum_{r=0}^{n} \frac{(-1)^{n+1} \, n^{2}}{2^{n}} \, \frac{(ax)^{2r}}{(2r)!} \\ &= \frac{2}{27} \, \cosh(ax) - \sum_{n,r=0}^{\infty} \frac{(-1)^{n+r+1} \, (n+r)^{2} \, (ax)^{2r}}{2^{n+r} \, (2r)!} \\ &= \frac{2}{27} \, \left[ \cosh(ax) - \sum_{r=0}^{\infty} \frac{(-1)^{r}}{(2r)!} \, \left(\frac{ax}{\sqrt{2}}\right)^{2r} \, (1 + 6 r - 9 r^{2}) \right] \\ &= \frac{2}{27} \, \left[ \cosh(ax) - \cos\left(\frac{ax}{\sqrt{2}}\right) - \frac{3 x}{4} \, \partial_{x} \left[ \cos\left(\frac{ax}{\sqrt{2}}\right) \right] + \frac{9 x^{2}}{4} \, \partial_{x}^{2} \left(\cos\left(\frac{ax}{\sqrt{2}}\right)\right) \right] \\ &= \frac{2}{27} \, \left[ \cosh(ax) - \frac{9 a^{2} \, x^{2} + 8}{8} \, \cos\left(\frac{ax}{\sqrt{2}}\right) + \frac{3 a \, x}{4 \, \sqrt{2}} \, \sin\left(\frac{ax}{\sqrt{2}}\right) \right] \end{align}