Mathematics is more than just solving problems. This means we should take care of the message of the problem rather than its numerical answer. A nice way (a personal favorite) to achieve this is to look at the problem in a reduced context. We can gain much more insight when we loosen some strong ideas underlying in the question. One is then recommended to solve the weaker problem (which is in most cases, more difficult to solve). This is the core of what most mathematicians do in their research. They generalize the already solved problem under a weaker assumption.

Why do we care then? This may have a variety of benefits. From my perspective, the importance in generalization lies in the applicability of the statement. If the statement is too strong, i.e. applicable in only the particular problem you are solving at the moment, it is not very useful; we wish to use the ‘idea’ (hence the term ‘message’ in paragraph 1) to solve other problems as well.

For example, a high schooler would consider the geometric sum $1/2+1/4+\cdots$ with a common ratio 1/2. A numerical answer would be then simply $S_\infty=1$. However, we cannot gain much more insight with this, as it only applies to this problem alone. Instead, the high-schooler can loosen the assumption on the common ratio and ask how the sum converges if $r=x$ . Then we may use the usual idea $S_\infty=\frac{1}{1-x}$ when we assume the initial term of the sequence is 1. This gives much more insight as we can gain knowledge of the infinite-degree polynomial $1+x+x^2+\cdots=\frac{1}{1-x}$. (of course, we will think in the region $\vert x \vert<1$ for convergence).

A motivated high-schooler then may think of the integral $\int\frac{1}{1-x}dx=-\ln\vert(1-x)\vert+C$, and realize we may rather integrate the power series on the left hand side to obtain $x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$, and obtain the peculiar identity

$$ \ln\vert(1-x)\vert=-(x+x^2/2+x^3/3+\cdots) $$

where we’ve fixed the integration constant C=0 for agreement at x=0. This type of ‘assumption-loosesning’ approach allows us to arrive at very interesting results. The above does not only restric to high-schoolers. A mathematician may now even drop the assumption $\vert x \vert<1$ and see what happens. Since the partial sum $1+x+x^2+\cdots+x^{n-1}=\frac{x^n-1}{x-1}$, one may note instead the factorization

$$ x^n-1=(x-1)(1+x+x^2+\cdots+x^n) $$

for any $x\in \mathbb{R}$. Now, the infinity $\infty$ is not regarded as a member of the real numbers, but the mathematician may choose to work out in the extended real line, $\mathbb{R}\cup{\infty}$, i.e. decide to treat infinity as a ‘number’. Let’s see what happens. This gives a really interesting result, as simply then

$x^{\infty}-1=(x-1)(1+x+x^2+\cdots)$. One might ask why this bizarre looking expression is even useful. However, we may substitute $x=-1$ in the above identity and obtain

$$ (-1)^\infty-1=(-2)\cdot(1-1+1-1+\cdots). $$

which suggests that the alternating series $1-1+1-\cdots$ can be realized as $\frac{1}{2}+\frac{(-1)^\infty}{2}.$ This means that the (usually known to be ‘divergent’) alternating series can be separated into a convergent part and a divergent part! This is very interesting. This method can be generalized further, noting

$$ 1+x+x^2+\cdots=\frac{x^\infty}{x-1}+\frac{1}{1-x}:=D(x)+C(x) $$

where $D(x)=x^\infty/(x-1)$ is the divergent part of the series and $C(x):=1/(1-x)$ is its convergent part. The mathematician then notes that if $\vert x \vert<1,$ the divergent part is no longer divergent and converges to zero, and hence leads to the usual geometric series formula.

The curious mathematician may differentiate the above identity in the extended real number system, leaving

$$ 1+2x+3x^2+\cdots=\frac{\infty x^{\infty-1}-x^\infty}{(x-1)^2}+\frac{1}{(1-x)^2}, $$

suggesting that the divergent part of this series is $\frac{\infty x^{\infty-1}-x^\infty}{(x-1)^2}$ and its convergent part $\frac{1}{(1-x)^2}$. If we substitute $x=-1$, then we obtain the divergent and convegent parts of the alternating series