Question

CALC 2

Answer 1

Can you please help me solve the first one ? https://www.dropbox.com/s/j0zdknkp2vi7y5f/Screenshot%202013-12-14%2015.25.04.jpg

Answer 2

@hero

Answer 3

@skullpatrol

Answer 4

@atlas

Answer 5

@AllTehMaffs

Answer 6

Can you tell if it's a geometric or arithmetic series?

Answer 7

Well, hint, I don't believe that any of the series in the set you gave are arithmetic.

So, in any case, since it's geometric, that means you can set it to the form
\[ \sum_{k=1}^{\infty} a(r)^{k-1}\]
and if r>1 it diverges

if r<1 it converges to the sum
\[ \sum_{k=1}^{\infty} a(r)^{k-1}=a\frac{1}{(1-r)}\]

Answer 8

How can you differentiate a geometric from an arithmetic serie @AllTehMaffs

Answer 9

An arithmetic sequence is where the difference between terms is the same, ie

1+3+5+7+9 etc...

They share a common difference.

A geometric term shares a common ratio, ie

\[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....\]
etc. They're all a term to some power.

This example is a geometric series, because it fits the latter form

\[1-\frac{3}{4}+\frac{9}{16}.....\]

So, looking at the series

\[ \sum_{k=1}^{\infty} \left( -\frac{3}{4} \right)^{k-1}\]

can you squish it into

\[ \sum a(r)^n\] ??
a=?
r=?
n=k-1 <---- this is trivial

Answer 10

wait "A geometric term shares a common ratio" , do you mean that A geometric series must have fractions ? basically what do you mean by "common ratio" ? which 2 numbers would have a common ratio ?

Answer 11

consider it as \[\sum_{a}^{b} =\int\limits_{a}^{b}dx\]

Answer 12

a sum can be considered as an integral as well ?? Also is that the reason this serie so considered "geometric"? http://screencast.com/t/xEyd3Qfftl7o

Answer 13

Throw out all those whose terms do not approach zero. 6 and 14 are Divergent.
Accept all that are obviously Geometric. 3, 4, and 12 Converge
Find those that are more subtly Geometric: 5 and 13 Converge
What's left?

Answer 14

I dont understand what meaning is underlined under the word "geometric"

Answer 15

I am pretty confused I guess

Answer 16

is this geometric because it has a power that includes this k variable?

Answer 17

So first, to make it less confusing, the k is just for counting, it's not an unknown. it's for distinguishing the individual terms when you write it out.

Second, it's a "geometric series" not because it's a fraction, but because you can multiply each successive term by a common factor (-3/4)

ie

\[ \sum_{k=1}^{\infty}1\left(-\frac{3}{4}\right)^{k-1}=1\left(-\frac{3}{4}\right)^0+1\left(-\frac{3}{4}\right)^1+1\left(-\frac{3}{4}\right)^2+1\left(-\frac{3}{4}\right)^3..... + \cancel{1\left(-\frac{3}{4}\right)^{\infty}}^0\]
\[=1+1\left(-\frac{3}{4}\right)+1\left(-\frac{3}{4}\right)\left(-\frac{3}{4}\right)+1\left(-\frac{3}{4}\right)\left(-\frac{3}{4}\right)\left(-\frac{3}{4}\right).......\]

So each term you add is the previous term multiplied by -3/4

Answer 18

Geometric sequence:
\[a_n=a_1r^{n-1}\]\(a_1\) = first term,
r = common ratio

Answer 19

so here the common r is -¾ or +¾ ?????

Answer 20

agh it has to be -3/4

Answer 21

aah*

Answer 22

I see

Answer 23

yup yup yup!
a=1
r=-3/4

Answer 24

since the negative is inside the exponential, it makes a big difference to keep that negative sign on the r - even powers give a positive number, and odd powers give a negative number.

Answer 25

so this converges

Answer 26

buts whats it's sum ?

Answer 27

yup yup yup.

The sum is given by

\[=\frac{a}{1-r}\]
super slick.

Answer 28

ok but hold on, is that some kind of a formula ??

Answer 29

yeah, that's just the formula for the sum of a converging infinite geometric sequence

Answer 30

if you have (for a=1) the
\[sum=s=1+r^1+r^2+r^3+....\] ,
then multiply the sum by r
\[r (sum) = rs=r^1+r^2+r^3+....\],
then
\[s-rs=1+(r-r)+(r^2-r^2)+(r^3-r^3)+......=1\]

because every term in s gets cancelled by a term in rs except for 1 (the first term)

following through gives
\[s(1-r)=1\]
\[s=\frac{1}{1-r}\]

Answer 31

thanks a lot

Answer 32

^_^ Glad I could help!

Answer 33

so I can use this exact thing basically for every geometrical infinite eerie ? https://www.dropbox.com/s/q0oywr1u24nkmpx/Screenshot%202013-12-14%2017.06.53.jpg

Answer 34

If it converges, most definitely. It's really, really cool I think!!

Answer 35

:D I see

Answer 36

^_^