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clearly $\sum_{n=1}^{\infty} n $ is a divergent series

but there is a way to show that $\sum_{n=1}^{\infty} n =-\frac1{12}$ (not really meaningful to me)

My problems are,

1. can we show $\sum_{n=1}^{\infty} n $ is equal to another real number different from $-\frac1{12}$ ?

2. If there is a way, by using these results can we prove the series $\sum_{n=1}^{\infty} n $ is divergent?(using the method of contradiction)

thanks.

Bad English
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    Every once in a while, this $-1/12$ thing keeps on popping up. What's up with that? Have you search MSE? It has been discussed more than once... – imranfat Feb 01 '18 at 17:26
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    Related – Antonio Vargas Feb 01 '18 at 17:27
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    @imranfat, according to me, this is not one of those stupid question. Read again, I think it is asking a deeper question, is it even possible to assign this sum any other value? – King Tut Feb 01 '18 at 17:27
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    No need to fight the windmills. This series is divergent, full stop. Something else, which via sloppy notation is also labelled $\sum_{n=1}^{\infty}n$, is equal to $-\frac{1}{12}$. –  Feb 01 '18 at 17:27
  • I am not claiming to be a stupid question, but it has been around (also on MSE) for quite some time. That's all...It think it is better to first explore MSE on such a topic to see if the answers satisfies before posting a new one. – imranfat Feb 01 '18 at 17:28
  • @KingTut The series diverges. It has no value and cannot be assigned a value. It is a notation that has been introduced by an indian mathematician but the analytical continuation of $\zeta$ has no longer things in common with its first " sum -related " expression. – Atmos Feb 01 '18 at 17:30
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    @atmos I never said the series converges. But I ask why only this particular value could be ever assigned to it (zeta function regularisation etc..) – King Tut Feb 01 '18 at 17:37
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    Also, wouldn't we contribute to the confusion if we were able to prove that "$\sum_{n=1}^{\infty}n$" (regardless of whether you are going to extend Riemann's zeta function, or one of Dirichlet's series, or something else) "converges" to something different than $-\frac{1}{12}$. In the best case, we would have yet another meaning of "$\sum_{n=1}^{\infty}n$", incompatible with the two(?) we already have. –  Feb 01 '18 at 17:38

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