Solving Problems Submitted to MAA Journals (Part 5a)

The following problem appeared in Volume 96, Issue 3 (2023) of Mathematics Magazine.

Evaluate the following sums in closed form:

\displaystyle \sum_{n=0}^\infty \left( \cos x - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} \dots + (-1)^{n-1} \frac{x^{2n}}{(2n)!} \right)

and

\displaystyle \sum_{n=0}^\infty \left( \sin x - x + \frac{x^3}{3!} - \frac{x^5}{5!} \dots + (-1)^{n-1} \frac{x^{2n+1}}{(2n+1)!} \right).

When I first read this problem, I immediately noticed that

\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \dots - (-1)^{n-1} \frac{x^{2n}}{(2n)!} \right)

is a Taylor polynomial of \cos x and

\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} \dots - (-1)^{n-1} \frac{x^{2n+1}}{(2n+1)!} \right)

is a Taylor polynomial of \sin x. In other words, the given expressions are the sums of the tail-sums of the Taylor series for \cos x and \sin x.

As usual when stumped, I used technology to guide me. Here’s the graph of the first sum, adding the first 50 terms.

I immediately notice that the function oscillates, which makes me suspect that the answer involves either \cos x or \sin x. I also notice that the sizes of oscillations increase as |x| increases, so that the answer should have the form g(x) \cos x or g(x) \sin x, where g is an increasing function. I also notice that the graph is symmetric about the origin, so that the function is even. I also notice that the graph passes through the origin.

So, taking all of that in, one of my first guesses was y = x \sin x, which is satisfies all of the above criteria.

That’s not it, but it’s not far off. The oscillations of my guess in orange are too big and they’re inverted from the actual graph in blue. After some guessing, I eventually landed on y = -\frac{1}{2} x \sin x.

That was a very good sign… the two graphs were pretty much on top of each other. That’s not a proof that -\frac{1}{2} x \sin x is the answer, of course, but it’s certainly a good indicator.

I didn’t have the same luck with the other sum; I could graph it but wasn’t able to just guess what the curve could be.

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