Proof that $\sum_{n=0}^{\infty}(-1)^{n} = \frac{1}{2}$. Is there any error?

Question

So, I proved that: $$\int f(\ln x)\ dx = x \sum_{n=0}^{\infty}(-1)^{n} f^{(n)}(\ln x) \ \ \ +\ \ C$$

where $f^{(n)}$ is the nth derivative of $f$.

if we let $f(x) = e^{x}$ then $f^{(n)}(x) = e^x$ as well, so:

$$\int e^{\ln x}\ dx = x \sum_{n=0}^{\infty}(-1)^{n} e^{\ln x} \ \ \ +\ \ C$$ giving us:

$$\int x\ dx = x^{2} \sum_{n=0}^{\infty}(-1)^{n} \ \ \ +\ \ C$$

if we differentiate both sides with respect to $x$ we end up with:

$$x=2x \sum_{n=0}^{\infty}(-1)^{n}$$ giving us:

$$\sum_{n=0}^{\infty}(-1)^{n} = \frac{1}{2}$$

Am I doing something wrong or is this valid? Because I've learned that that series is divergent, yet through this method I've proven it to be $\frac{1}{2}$ wich is also its Cesàro sum.

Answer 1

Always keep in mind that a proof that holds for any finite case need not hold in the infinite case. In the context of your proof, the following equality holds for any $n\in\mathbb{N}$: $$\int_a^xf\circ\log t\,dt=x\sum_{i=0}^n(-1)^if^{(i)}\circ\log x-a\sum_{i=0}^n(-1)^if^{(i)}\circ\log a+(-1)^n\int_a^xf^{(n+1)}\circ\log t\,dt,$$ which you can easily adapt your above proof to show via induction. However, this does not imply that the equality holds as $n\to\infty$, because $\infty$ is not a natural number - induction simply does not work in the infinite case. For your choice of $f=\exp$, we have on the right-hand side: $$\lim_{n\to\infty}\left[x\sum_{i=0}^n(-1)^i\exp x-a\sum_{i=0}^n(-1)^i\exp a+(-1)^n\int_a^xt\,dt\right]$$ but this limit does not exist for any value of $x$, since the sum $\sum_{i=0}^\infty(-1)^i$ diverges. Therefore, you can't extend this identity to the infinite case without checking that the sum on the right-hand side actually converges - otherwise the equality is meaningless.

However, when that series above does converge and the "remainder term" consisting of the integral on the right goes to zero, you can still use it to get some interesting results - for example, try taking $f:x\mapsto x^n$, and you can find a series expression for the following integral: $$\int_a^x\log^nt\,dt$$

Answer 2

By mathematical induction $$\int_a^x f(\ln t)\ dt = \sum_{n=0}^{N}(-1)^{n} [tf^{(n)}(\ln t)]_a^x+(-1)^{N+1}\int_a^x f^{(N+1)}(\ln t) dt$$ is true for any $N\in\mathbb{N}$ and $a\gt0.$

But passing from this finite sum to an infinite series is rather technical. One obvious condition that can make this happen is restrict $f$ to be a polynomial and, then carefully extend the result for power series via Taylor polynomials. This in return will capture functions like $\exp(x)$ in your construction.

Since $\lim_{t\to 0^+} t(\ln (t))^n=0$ for any $n,$ we let $a\to 0^+$ to make things bit simpler. But, we still need to prove the uniform convergence for the infinite series by carefully choosing the domain of $f^{(n)}\circ\ln.$

If it is the case, I guess, after some effort we would be able to use Abel summations to conclude the desire result.

Answer 3

Let $$ g_n(x)=\sum_{k=0}^n(-1)^kxf^{(k)}(\log(x))\tag1 $$ then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}g_n(x) &=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n(-1)^kxf^{(k)}(\log(x))\tag2\\ &=\sum_{k=0}^n(-1)^kf^{(k)}(\log(x))+\sum_{k=1}^{n+1}(-1)^{k-1}f^{(k)}(\log(x))\tag3\\[6pt] &=f(\log(x))+(-1)^nf^{(n+1)}(\log(x))\tag4 \end{align} $$ which gives $$ g_n(x)=\int f(\log(x))\,\mathrm{d}x+(-1)^n\int f^{(n+1)}(\log(x))\,\mathrm{d}x\tag5 $$ $g_n$ tends to a limit if $f^{(n+1)}$ vanishes uniformly on compact sets. In the example in the question, $f(x)=e^x$, it does not. Instead we get $$ g_n(x)=\frac{1+(-1)^n}2\,x^2+C_n\tag6 $$ Plugging $(6)$ into $(2)$ and dividing by $x$ gives $$ 1+(-1)^n=2\sum_{k=0}^n(-1)^k\tag7 $$ which is a true statement, but does not lead to the sum tending to $\frac12$.