زائد$1$ - tradução para Inglês
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زائد$1$ - tradução para Inglês

DIVERGENT SERIES
1+1+1+···; 1 + 1 + 1 + 1 + 1 + · · ·; 1 + 1 + 1 + 1 + · · ·; 1 + 1 + 1 + 1 + …; 1 + 1 + 1 + 1 + ...; Zeta(0)
  • alt=A graph showing a line that dips just below the ''y''-axis

زائد      
overbid
جمع         
  • التفاحات]] 3+2=5
  • شكل يبين كون 4 + 2 = 2 + 4
  • تمثيل بياني على شكل مستقيم لعملية الجمع 2 + 4 = 6. إزاحة بمقدار اثنين فإزاحة بمقدار أربعة تكافئ إزاحة بمقدار ستة.
  • حدود عملية الجمع جاءت على شكل عمودي. الخط الأفقي يكافئ علامة التساوي والعدد الذي أسفله هو ناتج العملية : 5 + 12 = 17
  • علامة زائد
عملية رياضيّة
الجمع; Addition; 1+1=2; 1 + 1 = 2; جمع (رياضيات)
band, bean, gathering, scratch together, combination, ingathering, collecting, cast, collection, ally, pickup, grouping, joining up, uniting, aggregate, connecting, summation, summing up, assemblage
جمع         
  • التفاحات]] 3+2=5
  • شكل يبين كون 4 + 2 = 2 + 4
  • تمثيل بياني على شكل مستقيم لعملية الجمع 2 + 4 = 6. إزاحة بمقدار اثنين فإزاحة بمقدار أربعة تكافئ إزاحة بمقدار ستة.
  • حدود عملية الجمع جاءت على شكل عمودي. الخط الأفقي يكافئ علامة التساوي والعدد الذي أسفله هو ناتج العملية : 5 + 12 = 17
  • علامة زائد
عملية رياضيّة
الجمع; Addition; 1+1=2; 1 + 1 = 2; جمع (رياضيات)

collecting

Definição

one
the upper limit of intoxication or exhaustion
after the second pint of gin, i was hard one-ing

Wikipédia

1 + 1 + 1 + 1 + ⋯

In mathematics, 1 + 1 + 1 + 1 + ⋯, also written n = 1 n 0 {\displaystyle \sum _{n=1}^{\infty }n^{0}} , n = 1 1 n {\displaystyle \sum _{n=1}^{\infty }1^{n}} , or simply n = 1 1 {\displaystyle \sum _{n=1}^{\infty }1} , is a divergent series, meaning that its sequence of partial sums does not converge to a limit in the real numbers. The sequence 1n can be thought of as a geometric series with the common ratio 1. Unlike other geometric series with rational ratio (except −1), it converges in neither the real numbers nor in the p-adic numbers for some p. In the context of the extended real number line

n = 1 1 = + , {\displaystyle \sum _{n=1}^{\infty }1=+\infty \,,}

since its sequence of partial sums increases monotonically without bound.

Where the sum of n0 occurs in physical applications, it may sometimes be interpreted by zeta function regularization, as the value at s = 0 of the Riemann zeta function:

ζ ( s ) = n = 1 1 n s = 1 1 2 1 s n = 1 ( 1 ) n + 1 n s . {\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1-2^{1-s}}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{s}}}\,.}

The two formulas given above are not valid at zero however, but the analytic continuation is.

ζ ( s ) = 2 s π s 1   sin ( π s 2 )   Γ ( 1 s )   ζ ( 1 s ) , {\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\ \sin \left({\frac {\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s)\!,}

Using this one gets (given that Γ(1) = 1),

ζ ( 0 ) = 1 π lim s 0   sin ( π s 2 )   ζ ( 1 s ) = 1 π lim s 0   ( π s 2 π 3 s 3 48 + . . . )   ( 1 s + . . . ) = 1 2 {\displaystyle \zeta (0)={\frac {1}{\pi }}\lim _{s\rightarrow 0}\ \sin \left({\frac {\pi s}{2}}\right)\ \zeta (1-s)={\frac {1}{\pi }}\lim _{s\rightarrow 0}\ \left({\frac {\pi s}{2}}-{\frac {\pi ^{3}s^{3}}{48}}+...\right)\ \left(-{\frac {1}{s}}+...\right)=-{\frac {1}{2}}}

where the power series expansion for ζ(s) about s = 1 follows because ζ(s) has a simple pole of residue one there. In this sense 1 + 1 + 1 + 1 + ⋯ = ζ(0) = −1/2.

Emilio Elizalde presents a comment from others about the series:

In a short period of less than a year, two distinguished physicists, A. Slavnov and F. Yndurain, gave seminars in Barcelona, about different subjects. It was remarkable that, in both presentations, at some point the speaker addressed the audience with these words: 'As everybody knows, 1 + 1 + 1 + ⋯ = −1/2.' Implying maybe: If you do not know this, it is no use to continue listening.