Question

For a given series (see attached) write out the first few terms and find the series sum

Answer 1

Would you mind walking me through the process anyway? I could use the help :)

Answer 2

and I got the wrong answer by typing that into the program

Answer 3

I gotta say, you're damned smart for 14

Answer 4

This is an infinite series...Use the infinite series formula for geometric sequences:\[\sum_{n=0}^{\infty}ar^n=a \left( \frac{ 1 }{ 1-r} \right) \]Here your a = 1, r = -1/10 and you can re-write the fraction -1^n/10^n as (-1/10)^n. So here your a = 1, r = -1/10, and since this is an infinite series, you use the formula by simply plugging in a = 1 and r = -1/10 and evaluate.\[\sum_{n=0}^{\infty}\frac{ -1^n }{ 10^n }=\sum_{n=0}^{\infty} \left( -\frac{ 1 }{ 10 } \right)^n=1\left( \frac{ 1 }{ 1-(-\frac{ 1 }{ 10 }) } \right)=\frac{ 1 }{ 1+\frac{ 1 }{ 10 } }=\frac{ 1 }{ \frac{ 11 }{ 10 } }=\frac{ 10 }{ 11 }\]And that's your answer.

@karama

Answer 5

It is correct -.-

Answer 6

the series is geometric the first few terms are

1, -1/10, 1/100, -1/1000...

so the 1st term is 1, the common ratio r = -1/10 so the series has a limiting sum since

|-1/10| < 1

the formula for a limiting sum is

\[s _{\infty} = \frac{a}{1 - r}\]

substitute your values for the 1st term a and the common ratio r to find the sum of the series.

Answer 7

it is not

Answer 8

so plug in values. Starting with n = 0.
So find WHEN n = 0
So find WHEN n = 1
So find WHEN n = 2
So find WHEN n = 3
So find WHEN n = 4
So find WHEN n = 5

These would be your first five terms.

Answer 9

your error appear to be the site seems to ask for the 1st term , which is 1... and you have entered the limiting sum 10/11

Answer 10

shouldn't my series converge to 0?

Answer 11

oh...my bad on that

Answer 12

nope.... it won't it converges to 10/11.

Answer 13

So I have listed the first few terms above.... hope it helps.

Answer 14

the series is

1 - 1/10 + 1/100 - 1/1000 + 1/10000

so it is impossible for it to get to zero... as you always subtract a smaller term from a larger term

Answer 15

yeah...I ended up plugging in the terms, but since this series has negatives and goes on infinitely, shouldn't the eventually add to 0?

Answer 16

and the abs(r) <1 doesn't that also support convergance

Answer 17

nope... look again at your terms

1st term 1

2nd term -1/10

so their sum is 9/10

3rd = 1/100

4th -1/1000

so the sum of 3rd and 4th is 9/1000

Answer 18

it does.... if -1 < r < 1 is the same as | r | < 1

so if the common ratio, r, lies between -1 and 1 the series will converge

Answer 19

thats why you can apply the limiting sum

Answer 20

So If \[\left| r \right| < 1 \] it converges.

Answer 21

And If \[\left| r \right| > 1\] it diverges

Answer 22

If it converges, then the sum converges to \[S_{n} = \frac{ a_{1} }{ 1 - r }\]

Answer 23

and my r is -1/10 correct?
because (-1)^n/10^2=(-1/10)^n is that right?

Answer 24

that is correct.

the common ratio is -1/10

Answer 25

and the 1st term in the series is 1

Answer 26

so then a1=1
r=-1/10
1/(1+1/10)=1/(11/10)=1*10/11=10/11

Answer 27

yes so now that you have r and the first term just plug them back into the formula to find where it converges to.

Answer 28

thats correct

Answer 29

and what is \[s_n\]

Answer 30

Its just the notation that I used to show SUM, but you don't have to worry about it.

Answer 31

cool

Answer 32

and this means the series converges to 10/11?

Answer 33

thats correct

Answer 34

Yep

Answer 35

success!

Answer 36

You can take a look and write down the 'formulas' I gave you previously if you think it might help you.

Answer 37

I absolutely will

Answer 38

I want to give you both best responses

Answer 39

lol... all you needed to do was read the screen questions... people had given you the answers yesterday

Answer 40

^.^
GOOD JOB!!

Answer 41

it was pretty darn late....I don't think I even knew what I was typing lol...and thank you both

Answer 42

No problem :D

Answer 43

glad to help

Answer 44

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