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By considering the partial sums for S, that is Sn =1+2+3+···n show that the infinite series S does not converge.

However in this video http://www.numberphile.com/videos/analytical_continuation1.html it states that this series converges to -1/12? How can this be?

Hanul Jeon
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oander66
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    That's what you call an "abuse of notation". –  May 10 '15 at 23:52
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    It doesn't converge to $-\dfrac1{12}$. It sums to $-\dfrac1{12}$… if you're willing to butcher the meaning of "sum". The normal meaning of "sum" (of an infinite series) is the limit of the partial sums, but that's clearly not what they're using there. When you try and assign a value to divergent sums, you're not actually summing them in any normal sense. (Even if the result is meaningful somehow, it's still a bit of an abuse of terminology to call it it's "sum".) – Akiva Weinberger May 11 '15 at 00:00
  • So is this video incorrect? @columbus8myhw – oander66 May 11 '15 at 01:14
  • @oander66 More or less. Mind you, it does add to $-\dfrac1{12}$ in a sense, but in a very convoluted sense that really isn't what you'd expect when they say that it's "what you get when you add them all together". (Besides: Call $S=1+2+3+4+5+\dotsb$. Let $S'=0+1+2+3+4+\dotsb$, and let $S''=0+0+1+2+3+\dotsb$. If they all had sums, we would clearly have $S=S'=S''$. Thus, we would have $0=S-2S'+S''$. But, if you actually add them up term-by-term, you'd find that $S-2S'+S''=1$. Contradiction.) – Akiva Weinberger May 11 '15 at 01:20
  • Did you search the site at all? To put it mildly, this is not the first tiime this question has been asked here :-) – Jyrki Lahtonen May 11 '15 at 05:27

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