S-1-i-1-i-1-1-i-i-S-i-i-1-i-1-i-1-S-i-i-S-S-1-iS-S-1-i-1-S-1-1-i-a-Is-this-correct-b-Do-there-exist-any-other-sequences-in-the-form-of-S-a-1-a-n-a-1-a-n

Question Number 14047 by FilupS last updated on 27/May/17

S = 1 + i - 1 - i + 1 + \dots

\frac{1}{i} = - i

S = i (- i + 1 + i - 1 - i + 1 + \dots)

S = i (- i + S)

S = 1 + i S

S (1 - i) = 1

∴ S = \frac{1}{1 - i}

a) Is this correct ?

b) Do there exist any other sequences

in the form of :

S = (a_{1} + \dots + a_{n}) + (a_{1} + \dots + a_{n}) + \dots

S = (a_{1} + \dots + a_{n}) (\underset{m times}{1 + 1 + \dots + 1})

\Rightarrow S = \underset{i = 1}{\sum^{m \to \infty}} \underset{j = 1}{\sum^{n}} a_{j}

where a_{t + 1} = {b a}_{t}, a_{1} = {b a}_{n}

I^{'} m very interested in these sequences

Commented by prakash jain last updated on 27/May/17

This is geometric series with

common ratio i .

Since ∣ i ∣= 1, by geometric series

test the series diverges .

S_{4 n} = 0

S_{4 n + 1} = 1

S_{4 n + 2} = 1 + i

S_{4 n + 3} = i

S = \frac{0 + (1) + (1 + i) + (i)}{4} = \frac{1 + i}{2}

Commented by prakash jain last updated on 27/May/17

S = \frac{1}{1 + i} is not correct . You can

get 4 different answers for S .

S = 1 + i - 1 - i + 1 + i - 1 - i

= (1 + i) - (1 + i) + (1 + i) + \dots

= (1 + i) - [(1 + i) - (1 + i) \dots]

= (1 + i) - S

2 S = (1 + i) \Rightarrow S = \frac{1 + i}{2}

Commented by RasheedSindhi last updated on 27/May/17

S = 1 + i S

\overset{?}{\Rightarrow} S (1 + i) = 1

Commented by ajfour last updated on 27/May/17

S_{n} = (\frac{1}{2} + \frac{i}{2}) (1 - i^{n})

S_{4 n} = 0

S_{4 n + 1} = 1

S_{4 n + 2} = 1 + i

S_{4 n + 3} = i .

Commented by ajfour last updated on 27/May/17

Commented by prakash jain last updated on 27/May/17

S_{n} is fine geometric series

and finite number of terms

converges .

Geometric series with ∣ r ∣⩾ 1

diverges .

For example

S = 1 - 1 + 1 - 1 + \dots

diverges and does not have a fixed

sum .

S = 1 o r 0

so S can be taken as \frac{1}{2} using

analytical continuation .

Commented by FilupS last updated on 27/May/17

S = \frac{1}{1 + i} has been corrected to S = \frac{1}{1 - i}

This is all interesting .

I am aware of S = 1 - 1 + 1 - 1 + \dots

These kinds if sequences are interesting

Commented by prakash jain last updated on 27/May/17

For the sequence of the type .

a_{t + 1} = {n a}_{t}, a_{1} = {b a}_{n}

a_{n} = a_{1} b^{n - 1}

a_{1} = {b a}_{n}

\Rightarrow b^{n} = 1

b^{n} = 1 = e^{2 π i}

b = e^{2 π k i / n}, k \in {0, 1, 2, \dots, n - 1}

b is n^{t h} root of unity .

Commented by ajfour last updated on 27/May/17

i f S = 1 - 1 + 1 - 1 + \dots

t h e n e v e n w e h a v e a f o r m u l a

f o r S_{n} :

S_{n} = \frac{1}{2} (1 - i^{2 n}) .

Commented by prakash jain last updated on 27/May/17

\frac{1}{1 - i} = \frac{1 + i}{2} is correct .

This value is correct with analytic

continuation . This series itself

is divergent .

S = 1 + 2 + 2^{2} + \dots .

S = 1 + 2 (1 + 2 + 2^{2} + . .)

S = - 1 .

S = - 1 is derived from analytical

continuation of f (x) = \frac{1}{1 - x}

= 1 + x + x^{2} + \dots

Commented by ajfour last updated on 27/May/17

i n c o m p r e h e n s i b l e i n d e e d!

Commented by prakash jain last updated on 27/May/17

Analytical continuation is interesting

for example using analytical

continuation it can be proved

that

1 + 2 + 3 + 4 + \dots to infinity = - \frac{1}{12}

Commented by FilupS last updated on 27/May/17

Yeah, i know a bit about the topic .

I {don}^{'} t really see^{'} {h o w}^{'} half of the solutions

are logical, but i love it

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