How Does the Triangle Inequality Help Solve a Geometric Series Problem?

[squaremeplz]

Homework Statement

Use the triangle inequality (many times) and the formula for the partial sums of a geometric series to show that for m>n

|s_m - s_n| <= r^(n-1)*(1/1-r)|s_2 - s_1|

Homework Equations

geometric series s = 1/1-r = 1 + r + r^2 + r^3...

The Attempt at a Solution

my first step was to multiply the terms inside the absolute to get 1/(1-r)

|s_m - s_n| <= r^(n-1)*|(s_2/1-r) - (s_1/1-r)|

next I expanded the geo. series as follows

|s_m - s_n| <= r^(n-1) *

|(s_2 + (s_2)*r + (s_2)*r^2 +...+ (s_2)*r^(m-1)) - |(s_1 + (s_1)*r + (s_1)*r^2 +...+(s_

1)*r^(m-1))|

then I applied the triangle inequality... many times

r^(n-1)*|(s_2 + (s_2)*r + (s_2)*r^2 +...+ (s_2)*r^(m-1)) - |(s_1 + (s_1)*r + (s_1)*r^2 +...+(s_1)*r^(m-1))|

<=

r^(n-1)*|(s_2 - s_1)| + r^(n)*|(s_2 - s_1)| + r^(n+1)*|(s_2 - s_1)| + ... + r^(m+n-2)*|(s_2 - s_1)|

is this close to the right path or did I make a mistake? thanks!

 

[HallsofIvy]

Homework Statement

Use the triangle inequality (many times) and the formula for the partial sums of a geometric series to show that for m>n

|s_m - s_n| <= r^(n-1)*(1/1-r)|s_2 - s_1|

Homework Equations

geometric series s = 1/1-r = 1 + r + r^2 + r^3...

An even more relevant equation is that s_m= 1+ r+ r^2+ ...+ r^m

The Attempt at a Solution

my first step was to multiply the terms inside the absolute to get 1/(1-r)

|s_m - s_n| <= r^(n-1)*|(s_2/1-r) - (s_1/1-r)|

But you don't know this is true. This is what you WANT to prove.

next I expanded the geo. series as follows

|s_m - s_n| <= r^(n-1) *

|(s_2 + (s_2)*r + (s_2)*r^2 +...+ (s_2)*r^(m-1)) - |(s_1 + (s_1)*r + (s_1)*r^2 +...+(s_

1)*r^(m-1))|

then I applied the triangle inequality... many times

r^(n-1)*|(s_2 + (s_2)*r + (s_2)*r^2 +...+ (s_2)*r^(m-1)) - |(s_1 + (s_1)*r + (s_1)*r^2 +...+(s_1)*r^(m-1))|

<=

r^(n-1)*|(s_2 - s_1)| + r^(n)*|(s_2 - s_1)| + r^(n+1)*|(s_2 - s_1)| + ... + r^(m+n-2)*|(s_2 - s_1)|

is this close to the right path or did I make a mistake? thanks!

I think you are going on the right path- but backwards!

I would do this: |s_m- s_n|\le |s_m- s_(m-1)|+ |s_(m-1)+ s_n|\le |s_m- s_(m-1)|+ |s_(m-1)- s_(m-2)|+ |s_(m-2)- s_n| etc. until you have steps of 1 from s_m to s_n. Then use the fact that s_(k+1)- s_k= r^k so you have a sum of r^k from r^n to r^m. Then factor out r^n.

 

[squaremeplz]

|s_m- s_n|\le |s_m- s_(m-1)|+ |s_(m-1)+ s_n|\le |s_m- s_(m-1)|+ |s_(m-1)- s_(m-2)|+ |s_(m-2)- s_n| etc. until you have steps of 1 from s_m to s_n. Then use the fact that s_(k+1)- s_k= r^k so you have a sum of r^k from r^n to r^m. Then factor out r^n.

 

[squaremeplz]

thank you for your time. When i use the fact that

s_(k+1) + s_k = r^k

|s_m - s_n|&lt;=r^(^n^-^1)(|s_(_m_) - s_(_m_-1)|+|s_(m-1) - s_(m-2)|+...

then the term on the left hand side in the inequalities (i.es_m in |s_(m) - s_(m-1)|)<br />
is the first term in an infinite series and the second one is also the first term in another infinite series.

therefore the sum of their infinite differences is1/(1-r) * |s_2 + s_1|[\itex]<br /> <br /> did I factor out the n correctly? i also still don&#039;t know why we have s_2 and s_1 <br /> <br /> <br /> and they are 1 step apart<br /> <br /> does this sound right? thanks!

 

[HallsofIvy]

Well, you need to practice putting { } around subscripts and superscripts but it looks like you have figured out this problem!