Summation of a series with 2 different methods gives 2 different answers The objective here is to find the value of $S$, where $S$ is given by,
$$S = 1-{1\over2}+{1\over3}-{1\over4}+...$$
I did this using two methods, but both the methods give different answers.
Method 1:
Let
$$S' = 1+{1\over2}+{1\over3}+{1\over4}+...$$
Now,
$1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+...$
$\;\;\;\;-1\;\;\;\;\;\;\;-{1\over2}\;\;\;\;\;\;\;-{1\over3}\;\;\;...$
$\frac{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}{}$
$1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+...$
So,
$$S = S'-S' = 0$$
Method 2:
Using the formula for infinite Geometric Progression, we can say
$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$
Integrating both sides with respect to $x$ from $0$ to $t$ we get,
$$[x+{x^2\over2}+{x^3\over3}+{x^4\over4}+...]^t_0 = [-ln(1-x)]^t_0$$
$$t+{t^2\over2}+{t^3\over3}+{t^4\over4}+... = -ln(1-t)$$
Putting $t = -1$,
$$-1+{1^2\over2}-{1^3\over3}+{1^4\over4}-... = -ln(2)$$
$$1-{1\over2}+{1\over3}-{1\over4}+... = ln(2)$$
So,
$$S=ln(2)$$
After some searching I found that the 2nd method gives the correct answer. But why is the 1st method wrong? 
 A: The series you denote by $S'$ is the divergent harmonic series. You're implicitly using the theorem that the limit of the sum (or difference) of two sequences is the sum (or difference) of their limits. Since $S'$ has no limit, this theorem doesn't apply.
A: There is no single correct value. This conditionally convergent series can be rearranged and summed to any value, see the example in
https://en.wikipedia.org/wiki/Riemann_series_theorem 
A: Furthermore, Riemann's rearrangement theorem asserts that for any conditionally , but not absolutely, convergent  series $u_n$, there is a rearrangement $u_{\sigma(n)}$ of its terms such that the re-arranged series converges to any prescribed number, or $+\infty$ or $-\infty$.
A: I want to add a (minor?) point to the other answers.  You considered ∑aₙ and ∑(-1)ⁿaₙ in the case aₙ=n⁻¹.  What happens when one replaces this with aₙ=nˢ?
Answer: For any s except s=‒1 your argument “would work”.
(Quotes are needed since I assume a particular way of regularization of divergent series for s≥‒1.  However, I never saw any other way meaningfully used in the case of power-of-n series.)
Indeed, for s<‒1 your approach works due to the convergence of ∑aₙ.  For s>‒1, both series may be regularized by Dirichlet method; the key observation is that Dirichlet regularization is preserved by dilution (which replaces a₁+a₂+… by 0+a₁+0+a₂+0+…).  (Note that the alternating series is also Chesaro/Abel/Borel summable — to the same answer as by Dirichlet.  The Chesaro summability is almost completely elementary.)
However, for s=‒1 the Dirichlet regularization gives ∞ (due to a pole of ζ(z) when z=1).  So in your calculation you get ∞‒∞, and you see that it is not necessarily 0.
Essentially, what you demonstrated is that when one gets ∞ with an “advanced” summation method, one should be as careful as with the naive summation method.
