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In infinite series we say that some series coverage and some diverge but the series. 1 + 1/2 + 1/3 + 1/4 •••• why we say that it diverges and has no finite answer , as it looks converging to me and also that 1+2+3+4•••••. Is also diverging but has finite answer

Chander Keerti Katal
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    Divergence of $\sum\frac{1}{n}$: https://math.stackexchange.com/questions/5115/why-does-1-x-diverge – scoopfaze Jul 05 '19 at 20:54
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    Divergent series (such as all of yours) have no finite answers. See the mathematical topic "Divergent Series" for information on what we can do anyway. https://en.wikipedia.org/wiki/Divergent_series – GEdgar Jul 05 '19 at 20:55
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    The divergence of $$1+2+3+4+\cdots$$ and of $$1+1+1+\cdots$$ is obvious. It is a bit harder to show that $$1+1/2+1/3+1/4+\cdots$$ also diverges. Don't fall into the trap to think that $$1+2+3+4+\dots =-1/12$$ is valid just because of a sloppy formulation !! – Peter Jul 05 '19 at 21:09
  • @Peter Definitions of infinite sums other than the limit of partial sums are not sloppy, they're just answering different questions. However, many accounts of these definitions, or of results that follow from them without even explaining what the definitions are, are sloppy. – J.G. Jul 05 '19 at 21:11
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    The questioner should clarify whether he/she has this extension in mind. No matter how popular this example to make something nonsensical meaningful is, I do not agree with it because we surely can find another "method" leading to another value. – Peter Jul 05 '19 at 21:23
  • @Peter Any method that gives a finite value for $\zeta(-1)$ gives $-1/12$. There's nothing "nonsensical" about summability methods you haven't studied yet. The OP is asking to learn about such extensions, so they can't be expected to say, "is this the one people mean"? – J.G. Jul 06 '19 at 06:53
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    @J.G., it's not at all clear to me what OP is asking to learn. If the harmonic series "looks converging" to OP, then I doubt OP is ready to learn about methods that give finite values to $\zeta(-1)$. – Gerry Myerson Jul 06 '19 at 07:35
  • @J.G. My final comment : The zeta-function does not converge at $-1$, hence is not defined there. To do as it would converge and that in this case, it would give the value -1/12 , does not mean that it actually makes sense to give the sum this value. And even if we accept this method, how is this method superior to other methods we could use to give 1+2+3+... a value ? – Peter Jul 06 '19 at 11:38
  • @Peter Superiority of a definition is relative to the application, the context. Virtually all research on the Riemann zeta function concerns the analytic continuation for which we absolutely would write $\zeta (-1)=-\frac{1}{12}$. You'll also find a finite $\zeta(-3)$ very convenient to predict the strength of the Casimir effect. If you genuinely believe someone has incorrectly implied partial sums give a value they don't, you can call that out; but when a source makes it clear they're discussing something else, there's no issue. – J.G. Jul 06 '19 at 12:01

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1 + 1/2 + 1/3 + 1/4 •••• why we say that it diverges and has no finite answer , as it looks converging to me

Just because it looks convergent doesn't mean that it is. To see that it's divergent, just add some parentheses: $$(1)+({1\over 2})+({1\over 3}+{1\over 4})+({1\over 5}+{1\over 6}+{1\over 7}+{1\over 8})+...$$ It's easy to check that proceeding in this way, the harmonic series is the sum of infinitely many "blocks" each of which sums to $\ge{1\over 2}$, hence it has to diverge. It diverges very slowly, but it does diverge.

1+2+3+4•••••. Is also diverging but has finite answer

What? It pretty obviously does not have a finite answer. Maybe you're referring to the sometimes-encountered claim "$1+2+3+...=-{1\over 12}$," in which case it's important to point out that that's wildly misleading without context.

Noah Schweber
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There's more than one way to define the sum of infinitely many terms. Usually, mathematicians define it as the $n\to\infty$ limit of the sum of the first $n$ terms. In this sense, both series become infinite. (This is obvious for the integers; for the fractions, which form the so-called harmonic series, see here, or pick your favourite among many proofs).

A different definition of the sum is needed if you want $1+2+3+\cdots$ to be finite; you might, for example, use zeta function generalisation to get a value of $-\frac{1}{12}$. (Other methods will say the same, or that the sum is infinite.) By contrast, the harmonic series resists this make-it-finite method (but not all of them).

As for $1+1+1+\cdots$, a similar treatment to the above would obtain $-\frac12$; but, again, this isn't the usual partial-sums limit, which would of course be $\infty$.

Finally, other finite answers are possible with suitable summation methods, as @reuns has noted. So, by taking the right linear combination of two methods, we can get any answer we want. Therefore, we have to know which definition of an "infinite sum" we have in mind when we assert anything.

J.G.
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  • (-1) because of the "equation" $$1+2+3+\cdots=-\frac{1}{12}$$ This has appeared already far too often here. – Peter Jul 05 '19 at 21:04
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    @Peter I didn't type that equation. I explicitly said the partial sums diverge and a different notion of "summing" is needed to get a finite result, so I didn't peddle whatever misconception bothers you. – J.G. Jul 05 '19 at 21:10
  • Note with the regularization $f_n(t) = n^{-t^3} \cos(t \ln t)$ then $\overset{f_n}\sum \frac1n =\lim_{t \to 0} \sum_{n=1}^\infty f_n(t)\frac{1}{n} = \lim_{t \to 0} \frac{\zeta(1+it+t^3)+\zeta(1-it+t^3)}{2} = \gamma$. And you won't get either $\infty$ or $-1/12$ for every summation method applied to $\sum n$ – reuns Jul 08 '19 at 06:25
  • @reuns Thanks; fixed. – J.G. Jul 08 '19 at 07:53
  • We're both too late to edit earlier comments, but the $\ln t$ should be $\ln n$. – J.G. Jul 08 '19 at 07:59