Understanding summation of infinite series by defining a new function

Question

So, we've been working on the topic of infinite series at uni. We've learnt the basics, then the convergence criteria, and how to get the taylor expansion of some basic functions like $e^x$ and $\ln(x+1)$.

After that we dealt with finding the sum of a specific series. If you had a series $\sum{a_n}$ you would take a part of it and put it as $x$ and then define a function in that way. For example:

For $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}$$ the defined function was (seemingly defined out of nowhere) $f:[-1,1]\rightarrow\mathbb{R}$ $$f(x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}$$ and we noticed that the second derivative of the given function looked nicely. Then we applied some basic integral identities, and basically from there on it was understandable.$\\$

The next example was $$\sum_{n=0}^{\infty}{\frac{(-1)^n\cdot n}{(2n+1)!}}$$ and again the beginning looked like this: we define a function $f:\mathbb{R}\rightarrow\mathbb{R}$ as $$f(x)=\sum_{n=0}^{\infty}{\frac{(-1)^n\cdot n}{(2n+1)!} \cdot x^{2n+1}}$$ and then through totally random number manipulation we achieved $f(1) = (\cos{1} - \sin{1})/2$ and then of course it's understandable since the Taylor expansions of $\sin$ and $\cos$ are available to us even during the tests.$\\$

Now, I'm having a really hard time figuring out how this works, these examples were the only ones shown to us, and I've unsuccessfully tried solving the problems left to us for practicing this method. How in the world do you come up with a function like that? I've looked at these two examples and I pretty much understand them, I've tried to reconstruct them on my own, but I've only succeeded because I remember specifically which function to define. $\\$ I would appreciate if someone helped me define a function in any of the following problems, I would be very grateful. I hope this question won't be put on hold because I haven't said what I've been trying, because what I'm missing is the beginning of the problem-solving. If I even remotely knew which function to take it would be really helpful.

$a)$ $\sum_{n=2}^{\infty}{\frac{(-1)^n}{n^2+n-2}}$

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

$c)$ $\sum_{n=0}^{\infty}{\frac{(n+1)(n+2)(n+3)}{2^n}}$

Answer 1

If you are told to multiply two given matrices you just do it, and if you are told to differentiate a complicated expression $\Psi(x)$, containing $\exp$, $\log$, $\sqrt{\cdot}$, etc., with respect to $x$ you just do it. This is because in such cases you have learnt an algorithm that produces the result in a finite number of steps.

Now for a series $s:=\sum_{n=0}^\infty a_n$, where the $a_n$ are defined as hopefully simple functions of $n$ there is no such algorithm. In this context it is a radically new idea to embed the given unsolvable problem into a larger environment by considering the function $$s(x):=\sum_{n=0}^\infty a_nx^n$$ instead, and letting $x:=1$ in the end. In this realm new facts become available, and new operations are possible. For instance, $$\sum_{n=0}^\infty x^n={1\over1-x}\quad(|x|<1),\qquad \sum_{k=0}^\infty{x^n\over n!}=e^x\quad(x\in{\mathbb C})\ ,$$ etcetera, and such series can be differentiated or integrated termwise. All of this does not provide you with clear cut algorithms and solution schemes, but it opens up a wide field where you can try tricks and gain experience. Good luck!

A hint: In example c) you should look at the third derivative of the function $$f(x):=\sum_{n=0}^\infty{x^{n+3}\over 2^n}=8x^3\sum_{n=0}^\infty\left({x\over2}\right)^n\ .$$

Answer 2

$$ \begin{array} \text{a)} \sum_{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}= \sum_{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}\to \sum_{n=2}^{\infty}\frac{x^{n+2}}{(n+2)(n-1)} \\ \text{b)} \sum_{n=0}^{\infty}\frac{(-1)^{n+1}\cdot (2n+1)^3}{(2n+1)!}\to x^2\sum_{n=0}^{\infty}\frac{x^{n-1}\cdot (2n+1)^3}{(2n+1)!}\\ \text{c)} \sum_{n=0}^{\infty}\frac{(n+1)(n+2)(n+3)}{2^n}\to \sum_{n=0}^{\infty}(n+1)(n+2)(n+3)x^n \end{array} $$ I could elaborate on the processes used for each, however for now I only am giving the generating functions that I would use to solve each. I will say in working them through, they used the techniques in the examples, but in sometimes less obvious ways.

Answer 3

For the functions suggested in the proposed problem consider the following functional forms that can be manipulated into the results desired.

$a)$ $$f(x) = \sum_{n=2}^{\infty}{\frac{x^n}{n^2+n-2}}$$ or $$g(x) = \sum_{n=2}^{\infty} \frac{x^{n+1/2}}{n^2 + n -2}$$

For $f(x)$ it is fairly evident that it satisfies the differential equation $$ x^2 \, f'' + 2 \, x \, f' - 2 \, f = \frac{x^2}{1-x},$$ with $f(0) = 0$ and $f'(0)=0$ leads to a solution of the form $$\sum_{n=2}^{\infty}{\frac{x^n}{n^2+n-2}} = \frac{x}{9} + \frac{1}{6} + \frac{1}{3 \, x} + \frac{1-x^3}{3 \, x^2} \, \ln(1-x).$$ For the case of $x=-1$ then $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2} = \frac{12 \, \ln(2) - 5}{18}$$ For the case of $x = 1/2$ then $$\sum_{n=2}^{\infty} \frac{1}{2^{n} \, (n^2+n-2)} = \frac{16 - 21 \, \ln(2)}{18}.$$

A similar differential equation can be obtained for $g(x)$.

$b)$ $$f(x) = \sum_{n=0}^{\infty}{\frac{(-1)^{n+1} \, x^{2n+1}}{(2n+1)!}} = - \sin(x)$$

For this case consider the differential operator $\delta = x D$ for which $$\delta^3 f = - \delta^3 (\sin(x))$$ which leads to the series $$\sum_{n=0}^{\infty}{\frac{(-1)^{n+1} \, (2n+1)^3 \, x^{2n+1}}{(2n+1)!}} = x(1-x^2) \, \cos(x) - 3 \, x^2 \, \sin(x).$$ Consider when $x=1$ for which $$\sum_{n=0}^{\infty}{\frac{(-1)^{n} \, (2n+1)^2 }{(2n)!}} = 3 \, \sin(1)$$ and if $x=\pi$ then $$\sum_{n=0}^{\infty}{\frac{(-1)^{n} \, (2n+1)^2 \, \pi^{2n}}{(2n)!}} = 1-\pi^2.$$ Other values can be obtained by varying the process.

$c)$ $$f(x) = \sum_{n=0}^{\infty} x^{n+3} = \frac{x^3}{1-x}.$$ Differentiate this function three times to obtain $$\sum_{n=0}^{\infty} (n+1)(n+2)(n+3) \, x^n = \frac{6}{(1-x)^4}$$ such that $$\sum_{n=0}^{\infty} \frac{(n+1)(n+2)(n+3)}{2^{n}} = 96.$$

Similar methods and starting functions may be chosen to lead to the same end result, but may vary in the process.