Question

Confused how to go about this problem. Any help is appreciated

Answer 1

http://prntscr.com/awfv3n

Answer 2

you need two things for the alternating series test
the absolute value of the terms are decreasing
i.e without the alternating business
and also that the terms go to zero

Answer 3

pretty clear how to write this is sigma notation with a general term \(a_n\) yes? (or no?)

Answer 4

Not really. I understand the conditions of the alternating series test but i never really learned sigma notation

Answer 5

ok what is the pattern then?

Answer 6

(n+1)^2

Answer 7

once we identify the patter, writing it as a sum is the easy part

Answer 8

(-1)(n+1)^2

Answer 9

forget the +1

Answer 10

oh yeah true

Answer 11

also, don't you need a fraction?

Answer 12

(-1)/(n)^2

Answer 13

yea that is all
the 2 is a constant, we can write that in front of the big ugly sigma \[2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\]

Answer 14

ok that is not exactly right, that one alternates wrong
no matter

Answer 15

yeah I realized that right when I saw yours

Answer 16

\[2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\]

Answer 17

oh nvm i thought you were talking about mine. Mine didnt alternate. I see what you mean now though

Answer 18

ok so you need to show two things \[a_{n+1}<a_n\] i.e. \[\frac{1}{(n+1)^2}<\frac{1}{n}\]

Answer 19

oops typo again

Answer 20

\[\frac{1}{(n+1)^2}<\frac{1}{n^2}\]

Answer 21

1/n^2 is bigger because the value of denominator would be bigger on the other one

Answer 22

which i would just say, since it is completely obvious

Answer 23

and also that \[\lim_{n\to\infty}\frac{1}{n^2}=0\] which is another "doh"

Answer 24

yeah so all 3 conditions are met the series is convergent

Answer 25

how do we estimate the error?

Answer 26

i have no idea

Answer 27

ok maybe a little idea
since it alternates, add up a few terms and see when you get two that are within \(.05\) of each other

Answer 28

Im supposed to justify this in an essay so I feel as if there is a method

Answer 29

when you add the terms go back and forth plus or minus, so you get closer with each term to the true sum
if two are within \(.05\) of each other then you will never get further away

Answer 30

@jim_thompson5910 ?

Answer 31

well kinda what is said is the reasoning
i am sure there is a more succinct way to put it, and i am sure it is in your text too

Answer 32

I believe the la grange error bound thing could be used but I am not too familiar with it

Answer 33

that would be more of a plug and chug although it makes sense to just keep running the sum until we approach .05

Answer 34

well there is a bit of something else going on here since it alternates
that would not work if it did not, since you would not know how far away the true sum is

Answer 35

and you are not approaching \(.05\) you are getting to within \(.05\) of the sum
here is an example
http://www.wolframalpha.com/input/?i=2%281-1%2F4%2B1%2F9-1%2F16%29-2%281-1%2F4%2B1%2F9-1%2F16%2B1%2F25%29

Answer 36

the next difference will be less than \(.08\)

Answer 37

oh ok

Answer 38

so we just want to be .05 off the true answer

Answer 39

im looking around trying to see how to do this

Answer 40

yeah cant figure it out

Answer 41

@ganeshie8 @jim_thompson5910 can either of you potentially help?

Answer 42

You may do the part b by using
\[|s-s_n| \le b_{n+1}\]

Answer 43

Here \(b_n = \dfrac{2}{n^2}\)

Answer 44

since you want that error to be less than 0.05, you need to solve
\[b_{n+1} \le 0.05\]

Answer 45

\[\dfrac{2}{(n+1)^2}\le 0.05\]

Answer 46

2/(n+1)^2 <_ .05

Answer 47

and then solve for n. Would that be the amount of terms needed in the sequence?

Answer 48

\[(n+1)^2 \ge \dfrac{2}{0.05}\]

Answer 49

\[(n+1)^2 \ge 40\]

Answer 50

\[n+1 \ge 6.32\]

Answer 51

\[n \ge 5.32\]

Answer 52

that means you need at least 6 terms to make the error in estimation less than 0.05

Answer 53

Thank you so much

Answer 54

That was way easier than i thought it to be

Answer 55

$$2-\frac24+\frac29-\frac2{16}+\dots=\sum_{n=1}^\infty (-1)^n\frac2{n^2}$$clearly \(2/n^2\to0\) as \(n\to\infty\) so it converges by the alternating series test. to determine the number of terms needed, we find $$\bigg|\sum_{n=k}^\infty (-1)^n\frac2{n^2}\bigg|< 0.05$$
since \(2/n^2\) decreases monotonically, it follows that we simply need to find the last term that contributes at most \(0.05\): $$\frac2{n^2}<0.05\\2<0.05n^2\\40<n^2\\n\ge 2\sqrt{10}\approx 6.3$$ so we need at least the first \(7\) terms

Answer 56

oops, that should read \((-1)^{n+1}\)

my above reasoning is flawed, though -- that is a sufficient but not necessary condition for the error to be at most \(0.05\)

Answer 57

Ahh right, 6 terms is also at most, not at least

Answer 58

actually it *is* at least 6 terms, it's just not a tight bound