Some background. I was exploring the series expansion for $\ln(1+\cos x)$ in an attempt to expand it (at least, initially!) up to the third non-zero term, and along the way I unexpectedly stumbled upon the infinite series $1-1+1-1+...$ and $1-2+3-4+5-6+7-...$. Then I thought to expand it instead via a different method, as I had divergent series as coefficients which obviously wouldn't have helped in determining the terms exactly. The coefficients via this method (as expected) came out as reals, and I was then, out of curiosity, tempted to equate these respectively, arriving at a very unexpected (in this context) result: $1-1+1-1+1-1+1-...=\frac{1}{2}$ and $1-2+3-4+5-6+...=\frac{1}{4}$. What caught me as particularly surprising is that these are well-known results for the sums via other methodologies. The "proof" is below.
By Taylor's theorem, substituting the series for $\ln(1+x)$ and $\cos x$:
\begin{align} \ln(1+\cos x) &=\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{n}\left({\sum_{k=0}^\infty \frac{(-1)^{k}x^{2k}}{(2k)!}}\right)^{n}\right)\\ &=\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{n}\left(1-\frac{x^2}2+\frac{x^4}{24}-\cdots\right)^n\right)\\ \end{align}
We are only concerned with the first 3 terms overall, and terms whose order is 6 or above cannot contribute towards the coefficients of the lower order terms (0,2,4 in this case). So consider $\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^n$, ignoring order 6 or higher terms, for $n\in\mathbb{Z_{>0}}$:
\begin{align} \left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^2 &=\ 1-\frac{2x^2}{2}+\frac{8x^4}{24}-\cdots\\ \left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^3 &=\ 1-\frac{3x^2}{2}+\frac{21x^4}{24}-\cdots\\ \left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^4 &=\ 1-\frac{4x^2}{2}+\frac{40x^4}{24}-\cdots\\ \end{align} and so on.
Conjecture that (from the patterns in the coefficients) $$\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^m=1-\frac{mx^2}{2}+\frac{am^2+bm+c}{24}x^4-\cdots$$ for some $a,b,c\in\mathbb{R}$.
We know some $x^4$ coefficients from the manual calculations, so we can solve for a, b, c (using m = 1, 2, 3):
$$a+b+c=1$$ $$4a+2b+c=8$$ $$9a+3b+c=21$$ $$\implies a=3,b=-2,c=0$$
$$\therefore\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^m=1-\frac{mx^2}{2}+\frac{3m^2-2m}{24}x^4-\cdots$$ for m = 1, 2, 3.
Assuming this as an inductive hypothesis for some $m\in\mathbb{Z_{>0}}$, we have:
\begin{align} \left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^{m+1} &=\left(1-\frac{x^2}{2}+\frac{x^4}{24})^m(1-\frac{x^2}{2}+\frac{x^4}{24}\right)\\ &=\left(1-\frac{mx^2}{2}+\frac{3m^2-2m}{24}x^4-\cdots\right)\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)\\ &=1-\frac{m+1}{2}x^2+\frac{3m^2-2m+6m+1}{24}x^4-\cdots\\ &=1-\frac{m+1}{2}x^2+\frac{3(m+1)^2-2(m+1)}{24}x^4-\cdots\\ \end{align}
$$\therefore\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)^m=1-\frac{mx^2}{2}+\frac{3m^2-2m}{24}x^4-\cdots\forall{m}\in\mathbb{Z_{>0}}$$ by mathematical induction.
Substituting this expansion back into the initial series expression, ignoring any terms in $x^6$ or higher (again because these terms will not contribute towards lower order coefficients):
\begin{align} \sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{n}(1-\frac{x^2}{2}+\frac{x^4}{24})^n\right) &=\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{n}\left(1-\frac{nx^2}{2}+\frac{3n^2-2n}{24}x^4\right)\right)\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}-\frac{x^2}{2}\sum_{n=1}^{\infty}(-1)^{n-1}+\frac{x^4}{24}\sum_{n=1}^{\infty}(3n-2)(-1)^{n-1}\\ &=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}-\frac{x^2}{2}\sum_{n=1}^{\infty}(-1)^{n-1}+\frac{x^4}{24}\left(3\sum_{n=1}^{\infty}n(-1)^{n-1}-2\sum_{n=1}^{\infty}(-1)^{n-1}\right)\\ \end{align}
Now we've expressed the first 3 terms with the coefficients as series, let's expand it in a different way.
\begin{align} \ln(1+\cos x) &=\ln(2+(\cos x-1))\\ &=\ln\left(2\left(1+\frac{1}{2}\left(\cos x-1\right)\right)\right)\\ &=\ln2+\ln\left(1+\frac{1}{2}(\cos x-1)\right)\\ &=\ln2+\sum_{n=1}^\infty \left(\frac{(-1)^{n-1}}{n}\left(\frac{1}{2}{\sum_{k=1}^\infty \frac{(-1)^{k}x^{2k}}{(2k)!}}\right)^n\right)\\ &=\ln2+\frac{1}{2}\sum_{n=1}^\infty \left(\frac{\left(\frac{-1}{2}\right)^{n-1}}{n}\left(-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\right)^n\right)\\ &=\ln2+\frac{1}{2}\left(\left(-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots\right)-\frac{1}{4}\left(-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\cdots\right)^2+\cdots\right)\\ \end{align}
Here, we can leave from consideration any of the order 6 or higher terms in the first "bracket", any of the order 4 or higher terms in the second "bracket" and all terms in any subsequent "bracket" for the purpose of our expansion (only looking for the first 3 terms). None of these will contribute towards the coefficients of the first few terms, as the power of the "bracket" is incrementing by 1 at each step. We get:
\begin{align} \ln(1+\cos x) &=\ln2+\frac{1}{2}\left(\left(-\frac{x^2}{2}+\frac{x^4}{24}\right)-\frac{1}{4}\left(-\frac{x^2}{2}\right)^2+\cdots\right)\\ &=\ln2-\frac{x^2}{4}-\frac{x^4}{96}+\cdots\\ \end{align}
Now that we have 2 different expressions for $\ln(1+\cos x)$, we can compare their coefficients:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}-\frac{x^2}{2}\sum_{n=1}^{\infty}(-1)^{n-1}+\frac{x^4}{24}(3\sum_{n=1}^{\infty}n(-1)^{n-1}-2\sum_{n=1}^{\infty}(-1)^{n-1})=\ln2-\frac{x^2}{4}-\frac{x^4}{96}$$
From this:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln2;$$ $$-\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}=-\frac{1}{4}\implies\sum_{n=1}^{\infty}(-1)^{n-1}=\frac{1}{2};$$ $$\frac{1}{24}\left(3\sum_{n=1}^{\infty}n(-1)^{n-1}-2\sum_{n=1}^{\infty}(-1)^{n-1}\right)=-\frac{1}{96}$$ $$\implies3\sum_{n=1}^{\infty}n(-1)^{n-1}-2\left(\frac{1}{2}\right)=-\frac{1}{4}\implies\sum_{n=1}^{\infty}n(-1)^{n-1}=\frac{1}{4}$$
So the reason for the post was, why is it that this arrives at said conclusions, and where do the flaws in the method lie? What's the catch? These are obviously quite unintuitive (but interesting, nonetheless) values for such series, arrived at in a completely unforeseen (to me, at least) way. Would highly appreciate any clarifications/explanations.
P.S - I'm an A-level/high-school student, so if there are obvious issues then apologies; I haven't done any analysis yet.
Using the Taylor series of $\cos$ means you do $\log(1+y)$ around $y=1$ since $\cos 0=1$.
In particular the series there is only conditionally convergent (and only for $y \le 1$) so you cannot exchange summation as a series. However if you argue that you consider only values of $y<1$ and let $y \to 1$ you rediscover well-known summability results.
By Littlewood-Hardy Tauberian theorem they would become actual convergence results if $na_n$ is bounded, and note that for the first one $na_n= \pm 1$ so you actually have a true convergence result, but for the other two $na_n= \pm n, \pm n^2$ respectively, so there you do not get convergence, just summability as expected.