Was Euler right?

We have a differential equation $$ y + y' = f(x) $$ and assume $f$ is infinitely differentiable. And we want to find particular solution. Then,I set $$ y_p = f(x)-f'(x)+f''(x)...., $$ i.e., $y_p=\sum_{i=0}^{\infty} (-1)^{n} f^{(n)}(x)$. It is easy to observe this formal sum is apparently a particular solution and it really works for all polynomial. For example, if $f(x) = x^3$ then $y_p = x^3 - 3x^2 + 6x - 6$ and it is true. But, when $f(x)=\sin x$ we have $$ y_p=(1-1+1-1....) \sin x - (1-1+1-1...) \cos x $$ since $(1-1+1-1..)=\sum_{i=0}^{\infty}(-1)^n$ is divergent we have no result. Up to here, there is nothing interesting. Euler thought that $\sum_{i=0}^{\infty}(-1)^n = 1/2$ since its result change $(1,0,1,0...)$ for finite sum (It is thought to be famous mistake of Euler. Of course, nobody blames him. In his time, convergency was not defined exactly). But, if take this sum as $1/2$ as Euler said $$ y_p = (\sin x - \cos x)/2 $$ and surprisingly, it is really the particular solution of $y+y'=\sin(x)$. Why does it work? Should we take this sum as $1/2$?

Solution 1:

We might justify as follows: If we replace $(-1)^n$ with $q^n$ where $q\approx -1$ but $|q|<1$, the geometric series converges to $\frac1{1-q}$ and all is good. However, if we plug $\sum q^nf^{(n)}(x)$ (where the convergence does not depend on $q$ alone, but in general also the derivatives might "explode") into the differential equation we only get an almost-solution. Or an exact solution to a slightly different equation: $$\tag1y'-qy=f(x).$$ Specifically with $f(x)=\sin(x)$ we can compute the corresponding series and obtain $$\tag2\frac{\sin x+q\cos x}{1+q^2}$$ as solution for $y'-qy=f(x)$. In this form, any special role of $q=-1$ is eliminated (and in fact this even works with $|q|>1$). In other words: Stretching the validity of limit beyond its "official" domain of convergence gives us a correct result here because the series was only a tool and the apparent singularity problems it has become invisible in the final results.

One may say that we did not answer the question: "What is $\sum(-1)^n$?" Instead we answered: "If $\sum(-1)^n$ has a value, what should it consistently be?" This leads to the investigation of summability methods such as Cesàro mean. (To quote from the linked Wikipedia page: "any summation method that possesses these properties and which assigns a finite value to the geometric series must assign this value.")