Question

why does normal algebra fail when applied to series?

Answer 1

What do yo mean?

Answer 2

I mean, that unless a series behaves itself; normal algebra like commutative property fails to work

Answer 3

Didn't bother Euler...:-)

Answer 4

Can you give me an example of what you mean perhaps?

Answer 5

1-1+1-1+1-1+..... is what i can recall from my readings last night

Answer 6

1/2

Answer 7

does this series converge or diverge or have a limit ....

Answer 8

No divergent....

Answer 9

convergent...

Answer 10

Its a telescoping sum

Answer 11

its divergent becasue it aint well behaved enough to use the commutative property

Answer 12

It's a half...

Answer 13

I don't know what you mean, individual terms can be commuted.

Answer 14

a half?

Answer 15

\[\sum_{k=1}^{\infty}(-1)^k\]

Answer 16

I though it diverges because sum alternate?

Answer 17

* oscillate

Answer 18

the sequence diverges so the series diverges ...

Answer 19

algebra only applies to equilities and not to limits

Answer 20

hold the phone. you need to write down precisely what
\[\sum_{k=1}^{\infty} (-1)^k\] means. it has a specific meaning.

Answer 21

addition is defined for a finite number of addends (whatever they are called) so the question is one about limits, not rules of algebra

Answer 22

http://en.wikipedia.org/wiki/Grandi%27s_series

Answer 23

\[\lim_{n->\infty}(-1)^n\] doesnt exist; it doesnt go to zero therefore the sequence diverges right?

Answer 24

depends on your definitions, really...

Answer 25

if a sequence diverges; the series diverges; if i recall my readings :)

Answer 26

Look up Cesaro sum...

Answer 27

not because it doesn't go to zero but because it has no limit

Answer 28

the sequence of partial sums is 1, 0 , 1 , 0 ... and this clearly does not approach one single number

Answer 29

if a sequence converges, the series is still undetermined; but has a better chance of converging :)

Answer 30

Ok, I will save you the trouble...

In other words, the Cesàro sum of an infinite series is the limit of the arithmetic mean (average) of the first n numbers in the series, as n goes to infinity.

Answer 31

amistre you are confusing two things. the sequence of partial sums is this
\[a_1, a_1+a_2, a_1+a_2+a_3 ...\]

Answer 32

A whole set of different rules apply to divergent series (outside the realms of analysis).

Answer 33

it is not the sequence of the terms, but the sequence of the partial sums that must converge. for example
\[\sum (\frac{1}{2})^k\] sequence of partial sums are
\[\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, ..\]

Answer 34

the sequence of partial sums for 1 - 1 + 1 - 1 + 1 - 1 +...
is
1, 0 , 1 , 0 , 1 , 0 , 1 , 0 , ...

Answer 35

and it is this sequence , the sequence of the partial sums, that has no limit

Answer 36

very good splaining :)

Answer 37

really? thnx!

Answer 38

http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7