Limit of a geometric series

[Adam85]

So I know that the following is true:

if |a| < 1 then
lim{n -> inf} sum{i = 0 to n} a^i = 1 / (1 - a)

Is the following true? If so, how do I prove it?

Given: lim {n -> inf} f(n) = a /\ forall n, |f(n)| < 1
Show: lim{n -> inf} sum{i = 0 to n} f(n)^i = 1 / (1 - a)

This isn't a homework problem, but it's something I got myself stuck on in a larger proof. If there is some additional property of f that is required to make this true, then please let me know. Also, if you find that this is untrue, please share a counterexample.

 

[JeffM]

So I know that the following is true:

if |a| < 1 then
lim{n -> inf} sum{i = 0 to n} a^i = 1 / (1 - a)

Is the following true? If so, how do I prove it?

Given: lim {n -> inf} f(n) = a /\ forall n, |f(n)| < 1
Show: lim{n -> inf} sum{i = 0 to n} f(n)^i = 1 / (1 - a)

This isn't a homework problem, but it's something I got myself stuck on in a larger proof. If there is some additional property of f that is required to make this true, then please let me know. Also, if you find that this is untrue, please share a counterexample.

\(\displaystyle Given\ f(n) = a\ and\ |a| < 1:\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\sum_{i=1}^n\left\{f(n)\right\}^i = \lim_{n \rightarrow \infty} \sum_{i=1}^na^i = \dfrac{1}{1 - a}.\)

Is that what you are asking?

 

[Adam85]

\(\displaystyle Given\ f(n) = a\ and\ |a| < 1:\)

\(\displaystyle \displaystyle \lim_{n \rightarrow \infty}\sum_{i=1}^n\left\{f(n)\right\}^i = \lim_{n \rightarrow \infty} \sum_{i=1}^na^i = \dfrac{1}{1 - a}.\)

Is that what you are asking?

Close.
\(\displaystyle Given \displaystyle\lim_{n \rightarrow \infty}\ f(n) = a\ and\ \) ...

The rest of what you wrote is correct. I also had a slightly stronger assumption than |a| < 1, but perhaps what you wrote is strong enough.

 

[DrPhil]

So I know that the following is true:

if |a| < 1 then
lim{n -> inf} sum{i = 0 to n} a^i = 1 / (1 - a)

Is the following true? If so, how do I prove it?

Given: lim {n -> inf} f(n) = a /\ forall n, |f(n)| < 1
Show: lim{n -> inf} sum{i = 0 to n} f(n)^i = 1 / (1 - a)

This isn't a homework problem, but it's something I got myself stuck on in a larger proof. If there is some additional property of f that is required to make this true, then please let me know. Also, if you find that this is untrue, please share a counterexample.

This looks to me like an expression of the "ratio test" for convergence of a series. The terms become a geometric series as n-->inf, but because you don't know what the early terms of the series are, the series is not itself geometric and the sum is not 1/(1 - a).

 

[Adam85]

This looks to me like an expression of the "ratio test" for convergence of a series. The terms become a geometric series as n-->inf, but because you don't know what the early terms of the series are, the series is not itself geometric and the sum is not 1/(1 - a).

Thanks for your response. This argument would make sense if I were trying to find:
lim{n -> inf} sum{i = 0 to n} f(i)^i

but since I am looking for:
lim{n -> inf} sum{i = 0 to n} f(n)^i

I should be able to conclude that thee early terms of the series are "close" to a^i (when n is sufficiently large). Did you misread the problem, or does your response apply to this problem and I am failing to understand (which is likely)?

 

[daon2]

I think the assumption is that \(\displaystyle |a|=|\lim_{n\to\infty}f(n)|<1\) and the goal is to show the sequence

\(\displaystyle \displaystyle S_n = \sum_{i=0}^nf(n)^i \)

converges to \(\displaystyle \dfrac{1}{1-a}\). If I were the OP I would consider the fact that \(\displaystyle 1+x+x^2+\cdots+x^n = \frac{x^{n+1}-1}{x-1}\).

 

[DrPhil]

So I know that the following is true:

if |a| < 1 then
lim{n -> inf} sum{i = 0 to n} a^i = 1 / (1 - a)

Is the following true? If so, how do I prove it?

Given: lim {n -> inf} f(n) = a /\ forall n, |f(n)| < 1
Show: lim{n -> inf} sum{i = 0 to n} f(n)^i = 1 / (1 - a)

This isn't a homework problem, but it's something I got myself stuck on in a larger proof. If there is some additional property of f that is required to make this true, then please let me know. Also, if you find that this is untrue, please share a counterexample.

Back to the original statement of the problem - now that I am convinced you really did write exactly what you meant!

TWO things are happening in the limit n-->inf:
First, it is Given that EVERY f(n) is approaching a,
and the corresponding term in the sum is approaching a^i
Further, the number of terms is becoming infinite, so the series is approaching an infinite geometric series.
Then (if |a|<1) the sum does indeed approach 1/(1 - a)

 

[JeffM]

Daon has twice now pointed out that I have been misreading the problem. So I have deleted my posts premised on what in hindsight is an inexplicable misreading.