1. ## Sequences and Series

Consider the series:

1+1/2+1/4+1/8....

How many terms of this series must be taken before the sum first exceeds 1999/1000?

What I did was first input 1999/1000 into the series equation, so

Sn= n/2 (2u1 + (n-1)d)
1999/1000 = n/2 (2 + (n-1)(1/2) )
1999/1000 = n (2+ (1/2)n - (1/2) )
0 = 1/2 n^2 +3/2 n - 1999/500

Then i used the quad formula and got

1.7, -4.7

... but its wrong lol =|

i'm not sure what im supposed to do

2. Originally Posted by juliak
Consider the series:

1+1/2+1/4+1/8....

How many terms of this series must be taken before the sum first exceeds 1999/1000?

What I did was first input 1999/1000 into the series equation, so

Sn= n/2 (2u1 + (n-1)d)
1999/1000 = n/2 (2 + (n-1)(1/2) )
1999/1000 = n (2+ (1/2)n - (1/2) )
0 = 1/2 n^2 +3/2 n - 1999/500

Then i used the quad formula and got

1.7, -4.7

... but its wrong lol =|

i'm not sure what im supposed to do
You are using the sum of an arithmetic sequence on a geometric sequence. Try doing the same but using

$S_n = \frac{a(1-r^n)}{1-r}$

Where a is the first term and r is the common ratio

3. Well, the series is a geometric progression and not arithmetic progression. Hence ur formula for sum is wrong.

The formula for sum of a geometric progression is:

$Sn = ar^((n-1))/(1-r^n)$

where a is in the initial term in the series
r is the ratio of the second term to first term or third term to second term and so on....(in this case it is 1/2)
n is the no. of terms reqd

The denominator depends on whether r>1 or r<1. In this case, r<1, hence the formula

Substituting for above parameters and taking the fact that it is just less than 1999/1000, u get,

$(1/2)^((n-1))/(1-(1/2)^n)<1999/1000$

Solve for n.

4. thank you!

I simplified down to

1.9995 < 1/2 ^n

how can i find the n from that

5. Originally Posted by juliak
thank you!

I simplified down to

1.9995 < 1/2 ^n

how can i find the n from that
$ln(1.9995) < -n\, ln(2)$

Note I said that $ln(0.5)^n = ln(2^{-1})^n = ln(2^{-n})$