In this proof of euler's summation,After derivation it said that : Second thing he assumed was: $$\sum_{k=0}^{\infty} (k+1)x^k=1-2+3-4+5-6+...=\frac{1}{4}$$ & I want to know how we can get to this . Thanks
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Just place $x=-1$ in the derivative equation... – N74 Dec 21 '16 at 18:44
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When people answer your questions, it really is polite to acknowledge them... – The Count Dec 23 '16 at 05:13
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Thanks for your answer.& sorry I saw it just now:) – ali yazdian Dec 23 '16 at 10:51
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The statement, which you quoted correctly, wasn't quite written correctly in the first place, at a glance. It should say that the sum, when $-1=x$, gives that series. Simply write $\sum_{k=0}^{\infty} (k+1)(-1)^k$ and for $k=0$, we get $1$. For $k=1$, we have $-2$, and so on. This was what N74's comment meant.
Beyond that, the series is divergent. We see that the partial sums are $1, -1, 2, -2, 3, -3$, etc. So when they say Euler "assumed", we really mean he did just that. There is no proof because it isn't correct. The sum does not have a value in the classical sense, but in a heuristic sense (Euler did a lot of this sort of hand-waving, if you ask me, and it is amazing he got away with it).
The Count
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