Question

k=1 sigma k tends to infinity (-1/2)^k how can i find sum of this series..?

Answer 1

\[\sum_{k=1}^\infty-\frac{1}{2}^k\]

Answer 2

(-1/2)^k bro this is my qus.?

Answer 3

yes, not sure how to do this though.

Answer 4

how can i find a sum of this sereies..?

Answer 5

I think I'd look at the sum of the negative terms and the sum of the positive terms separately.

Answer 6

i solve this as (-1/2)^2 ans (-1/2)^(k-1) this is look a like geometrec series

Answer 7

\[\sum_{k=1}^\infty-\frac{1}{2}^k=\sum_{k=1}^\infty\frac{1}{2^{2k}}-\sum_{k=0}^\infty\frac{1}{2^{2k+1}}\]

Answer 8

brohter whole power (-1/2)^(k) not only 1^k

Answer 9

I know, but this doesn't look as pretty:
\[\sum_{k=1}^\infty(-\frac{1}{2})^k=\sum_{k=1}^\infty\frac{1}{2^{2k}}-\sum_{k=0}^\infty\frac{1}{2^{2k+1}}\]

Answer 10

then how can i calculate the sum of this series .. which formula can help me solve this question.?

Answer 11

dear thomas can u explain now 1/2^(2k) ? why..?

Answer 12

So I'm splitting the series up in two part: the even terms k=2,4,6,... and the odd terms k=1,3,5,...

If I put 2k in that exponent there I get exactly the even terms. Similarly I use 2k+1 for the odd terms.

Answer 13

Does that make sense?

Answer 14

dear i can;t understand what u do now in this step plz explain now..

Answer 15

This is the series we're looking at: -1/2 + 1/4 -1/8 +1/16-1/32+...
I just group the positive terms (the even terms): (1/4+1/16+...)+(-1/2-1/8-1/32-...)

Answer 16

I did the same in this step, I just use the sigma notation:
\[\sum_{k=1}^\infty(-\frac{1}{2})^k=\sum_{k=1}^\infty\frac{1}{2^{2k}}-\sum_{k=0}^\infty\frac{1}{2^{2k+1}}\]

Answer 17

oho .. then dear how can i
calculate the sum of this series?

Answer 18

Let's rewrite this a bit:
\[\sum_{k=1}^\infty\frac{1}{2^{2k}}-\sum_{k=0}^\infty\frac{1}{2^{2k+1}}=\sum_{k=1}^\infty\frac{1}{4^k}-\frac{1}{2}\sum_{k=0}^\infty\frac{1}{4^k}\]
agree?

Answer 19

I just realized we're doing this harder than is necessary. We can continue with this, or do the easy way, whatever you like.

Answer 20

1/2(2k+1)= 1/4^(k) this step plz explian..?

Answer 21

Notice the 1/2 in front of the series.

so 1/2^(2k+1)=1/2*(1/2^(2k))=1/2*1/4^k

Answer 22

ok yup ..continue.

Answer 23

You might be familiar with this:\[\sum_{k=0}^\infty x^k=\frac{1}{1-x}\]

(We could have used this from the beginning (the easy way))

Answer 24

So we can use the here \[\sum_{k=1}^\infty \frac{1}{4^k}=\frac{4}{3}-1\]
-1 because we start from k=1 and not k=0, so we not compensate for the first term.

Answer 25

no dear i dont know 1/1-x ?

Answer 26

Well, you do now. It's a standard rule, one of the most important for series.

Answer 27

can u explain now.? dear which name of this series..?

Answer 28

It's the simplest form of a power series.

Answer 29

It's called a geometric series.

Answer 30

I realize my teaching skills aren't at their best at the moment.

Answer 31

no bro u r gud teacher..

Answer 32

So anyway let's that rule from the beginning. So x= -1/2
\[\sum_{k=0}^\infty (-\frac{1}{2})^k=\frac{1}{1--\frac{1}{2}}=\frac{2}{3}\]

Answer 33

but this is from k=0, so we still need to compensate, because the our series start from k=1.

Answer 34

brothe in geometric series |r|=1/2 ? not -1/2

Answer 35

A geometric series doesn't have to use 1/2 actually, it can use any number, (-1/2) ^k is also a geometric series.

Answer 36

i can't understant what u say

Answer 37

1/2 +1/4+1/8+... is a geometric series, but it's not the only one. Any series in which you get the next term by multiplying the previous term by constant is a geometric series. So we multiply by 1/2 each time in that last example. But we can also multiply each term by 1/4 for instance, to get 1 + 1/4+ 1/16 +... In our case we multiply by -1/2.

Answer 38

oho ok.. the sum 2/3 ?

Answer 39

That was starting from k=0 so still need to subtract the first term from that to get the series starting from k=1. The first term is 1, so the answer is 2/3-1=-1/3

Answer 40

so brother why are u break the quetion see the top?

Answer 41

what?

Answer 42

sigma 1/2^k - sigmat 1/2^(2k+1) why do we need this step

Answer 43

That was the hard way I went for first, you don't need to do that step.

Answer 44

(-1/2)^k in this qus r=-1/2 ? or a =1?

Answer 45

What do you mean by r and a?

Answer 46

geometriec series sum calculute formula a/(1-r) ?

Answer 47

oh, a=1, r=-1/2 indeed.

Answer 48

me right na.?

Answer 49

You were right.

Answer 50

brother last step can u explain now (2/3)-1 how.?

Answer 51

The 1/(1-x) rule I told you about only works if the series starts from k=0. Our series starts from k=1. So if I calculate the series starting from k=0 and then subtract the first term, we get what we want. 2/3 is the value of the series starting from k=0 and 1 iss the first term. So we get 2/3-1=-1/3

Answer 52

dear bro 1/1-x this is formula..?

Answer 53

\[\sum_{k=0}^\infty x^k=\frac{1}{1-x}\]

Answer 54

hmm thnks bro

Answer 55

You're welcome, I'm off, see ya.