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Let: $$ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... = A $$ Then: $$ \left(1 + \frac{1}{3} + \frac{1}{5} + ...\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ...\right) = A $$ $$ \left(1 + \frac{1}{3} + \frac{1}{5} + ...\right) + \frac{1}{2} \left(1 + \frac{1}{2} + \frac{1}{3} + ...\right) = A $$ $$ \left(1 + \frac{1}{3} + \frac{1}{5} + ...\right) + \frac{1}{2} A = A $$ $$ 1 + \frac{1}{3} + \frac{1}{5} + ... = \frac{1}{2} A $$ $$ 2 \left(1 + \frac{1}{3} + \frac{1}{5} + ...\right) = A $$ So: $$ 1 + \frac{1}{2} + \frac{1}{3} + ... = 2 \left(1 + \frac{1}{3} + \frac{1}{5} + ...\right) $$ $$ \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... = 1 + \frac{1}{3} + \frac{1}{5} + ... $$ Finally: $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... = 0$$ But we know that: $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... = \log(2)$$

Where is the mistake ?
Is it because I assumed that this series has some value $A$ ?
If the sum is summable to some value, then the final equation will be true, right?

Sumanta
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Michal
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    Your mistake is that you assumed the series ha a finite sum, when in fact ot doesn't. The harmonic series diverges. – Ishraaq Parvez Apr 03 '21 at 11:12
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    I am trying to think on it in a bit other way. For example, for a long time people thought that 1+1+1+1+1... is not summable, like other series, but it turned out that behind this sum there stays Zeta function and so the value is possible to be calculated. So what if playing with some diivergent series there stays some other function that allows to calculate value for it but the function that is behind the scene is not known yet. – Michal Apr 03 '21 at 11:18
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    Um... no. A divergent series is divergent, unless proven other wise. – Ishraaq Parvez Apr 03 '21 at 11:19
  • For every real number $r$ , we can rearrange the summands such that the sequence tends to $r$. This always happens in sequences that converge only conditionally, but not absolutely. – Peter Apr 03 '21 at 12:07
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    @IshraaqParvez What? Certainly $\sqrt{-1}$ exists; in fact the most common definition of complex numbers has $\sqrt{-1}=(0,1)$. – David C. Ullrich Apr 03 '21 at 12:35
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    That's my point - it does not EXIST, rather it is DEFINED by us to denote a certain expression. They're used for the DEFINITION of complex numbers, and complex numbers help - well, because, they have a lot of applications in various fields, from engineering to physics. – Ishraaq Parvez Apr 03 '21 at 12:45
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    $A$ in your question does not exist. You cannot do algebra with it. – K.defaoite Apr 14 '21 at 14:36
  • How to calculate the series $-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\frac{1}{10}...$? explains why your rearrangement is erroneous as well as how to obtain the correct answer. – user21820 May 22 '21 at 03:17

2 Answers2

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tl;dr Yes, in a sense the problem was assuming the harmonic series has a numerical value, and using $\infty - \infty = \log 2$ to conclude $0 = \log 2$.


I sense from the clarifying comments that the real question is not about this specific series, but about a larger issue: If we're given a formal infinite series, when can we assign a real value to the sum in a way that respects usual rules for manipulating series? And so, how do we know we can't assign a numerical value to the sum of the harmonic series?

What does seem clear (unfortunately) is that splitting a divergent formal infinite series into two (or finitely many) "subseries" and manipulating those is not on its own capable of yielding a definition for the sum of a formal series, because we have elementary theorems that at least one subseries diverges, so we learn nothing.

Consider Cesàro convergence: Given a sequence $(a_{k})$, we consider the sequence of arithmetic means $s_{n} = \frac{1}{n} \sum_{k=1}^{n} a_{k}$, and ask whether this converges. If $(s_{n}) \to s$, we say our formal infinite series converges in mean to $s$. This definition gives a unique value to some divergent formal series, e.g. the formal geometric series with ratio $-1$ converges in mean to $1/2$. Cesàro convergence is not a finite splitting: There's an averaging process in which the number of terms being averaged is unbounded.

Similarly, consider regularization by a convergent series of functions, such as the zeta function, or the formal geometric series $$ \sum_{k=0}^{\infty} x^{k} = \frac{1}{1 - x}. $$ We can define the value of the formal series on the left to be the rational expression on the right for all real (say) $x \neq 1$. With this definition the formal geometric series with ratio $-1$ "regularizes" to $1/2$. But again we've done something more than algebra: We've extended a single numerical series to a formal power series on an interval of numbers, summed the series where it converges ($|x| < 1$), and then used the fact that the sum makes sense outside $(-1, 1)$.

Do these examples mean "the geometric series with ratio $-1$ really is equal to $1/2$" in some philosophical sense? I'm not going to touch that question, but only point out that mathematically the question is not (in my view) well-posed, because before we ask mathematically, we have to specify how a value is to be assigned to an expression.

For ordinary infinite series, we choose to form a sequence of partial sums and take the limit. It's not (in my view) that "$\sum_{k=0}^{\infty} x^{k} = \frac{1}{1 - x}$ for $|x| < 1$ in some absolute sense", it's that we have a useful definition of equality that formally obeys familiar properties of real arithmetic, and it's therefore convenient to speak of the series on the left as having a numerical value.

Is there some universal definition that separates all formal (real, say) series into "truly summable" and "truly divergent"? Perhaps, but it's easy to define a family of series whose convergence is algorithmically undecidable (e.g., generate terms of $1$ for each non-terminal step in an algorithm, and $0$s after the algorithm terminates, so that the series converges if and only if the algorithm terminates). It looks to me that the detailed workings of such a "universal convergence test" are outside human comprehension, and are certainly uncomputable in a practical sense.

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Precisely. The harmonic series diverges. The partial sums grow to as large a number as you please. There is no such $A$.

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Ethan Bolker
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