Sequences-please help

[kaja_love]

1) Big, big problems with absolute value bars when trying to find such natural numbers N(a) where |-1 - (2-n)/n| < e

e = 1/100
lim[n->00](2-n)/n = -1

We could write it:

-1 - (2-n)/n < -1/100

...or:

|-1 - (2-n)/n| < 1/100

If we write it:

-1 - (2-n)/n < -1/100
-2/n < -1/100
2/n > 1/100
200 > n //wrong

Second way: I am not even shure how to do since I don't know what should happen to the equation when we remove absolute value bars.

|-1 - (2-n)/n| < 1/100

I need to learn how to do this both ways.

2) For what x will infinite geometric sequence 3 + 2/x + 4/(3x^2) converge? Since common ratio k has to be |k| < 1, I think k = |2/(3*x)| < 1

What would best method be? I broken the problem in two parts. First part:

. . .2/(3*x) > 0

...and in second:

. . .2/(3*x) < 0

First part:

2/(3*x) < 1
2/3 < x

Second part:

-1 < -2/(3*x) < 0
-2/(3*x) < -1
2/3 > x

3) And last question: How many terms must we add for geometric sequence to be 18,27... > 1000

. . .common ration k = 3/2

1000 < 12*(k^n - 1)/(k-1)
512/12 < (n * log3)/(n * log2)

log512 < n1 * log3
log512/log3 < n

log12 < n * log2
log12/log2 < n2

n1 is not equal to n2 so wrong!

Thank you.

 

[soroban]

Hello, kaja_love!

Here's the last one . . .

How many terms must we add for geometric sequence 18, 27... to be greater than 1000 ?

Sum of the first \(\displaystyle n\) terms of a geometric sequence: \(\displaystyle \,S_n\:=\:a_1\cdot\L\frac{r^n\,-\,1}{r\,-\,1}\)
\(\displaystyle \;\;\)where \(\displaystyle a_1\) is the first term and \(\displaystyle r\) is the common ratio.

We have: \(\displaystyle \,a_1\,=\,18,\;r\,=\,\frac{3}{2}\)

Hence: \(\displaystyle \L\,18\cdot\frac{\left(\frac{3}{2}\right)^n\,-\,1}{\frac{3}{2}\,-\,1} \:>\:1000\;\;\Rightarrow\;\;36\left[\left(\frac{3}{2}\right)^n\,-\,1\right]\:>\:1000\)

\(\displaystyle \L\;\;\Rightarrow\;\;\left(\frac{3}{2}\right)^n\,-\,1\:=\:\frac{250}{9}\;\;\Rightarrow\;\;\left(\frac{3}{2}\right)^n\:>\:\frac{259}{9}\)


Take logs: \(\displaystyle \L\:\log\left(\frac{3}{2}\right)^n\:>\:\log\left(\frac{259}{9}\right)\;\;\Rightarrow\;\;n\cdot\log\left(\frac{3}{2}\right)\:>\:\log\left(\frac{259}{9}\right)\)

Hence: \(\displaystyle \L\:n\:>\:\frac{\log\left(\frac{259}{9}\right)}{\log\left(\frac{3}{2}\right)} \;=\;8.285801706\)

Therefore, \(\displaystyle n\,=\,9\) . . . We must add the first nine terms.


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check
\(\displaystyle n\,=\,8:\;\;\L S_8\:=\:18\cdot\frac{\left(\frac{3}{2}\right)^8\,-\,1}{\frac{3}{2}\,-\,1}\:=\:886.640625\;\) . . . under a thousand

\(\displaystyle n\,=\,9:\;\;\L S_9\;=\;18\cdot\frac{\left(\frac{3}{2}\right)^9\,-\,1}{\frac{3}{2}\,-\,1}\;=\;1347.960938\;\) . . . over a thousand

We're golden!

 

[Gene]

You are so close, but you removed the absolute value at the wrong time.
|-1-(2-n)/n| =
|-1-2/n+1| =
|-2/n| =
2/n
Now you have
2/n < 1/100
200<n
n>200

 

[kaja_love]

You are so close, but you removed the absolute value at the wrong time.
|-1-(2-n)/n| =
|-1-2/n+1| =
|-2/n| =
2/n
Now you have
2/n < 1/100
200<n
n>200

I'm really, really having problems understanding this. Why can't we do it

-1 - (2-n)/n < -1/100
          -2/n < -1/100  
            2/n > 1/100 // we multiply both sides with (-1) and '<' 
                                 becomes '>'. I followed the rule which 
                                 says you have to change signs if you 
                                 multiply with negative numbers, so why
                                 is result wrong ?
           200 > n

 

[Gene]

Look at your first step
|-1 - (2-n)/n| < 1/100
If n=2 that becomes
|-1| < 1/100
1 < 1/100
False

-1 - (2-n)/n < -1/100
becomes
-1 < -1/100
True.

I don't follow where you are getting -1/100 but this counter-example shows that it is not valid. They would have the same result if it were.