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April 9th, 2019, 01:06 AM   #1
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Limit of a series

Does anyone know how to solve this limit?

$$\lim_{n\to\infty} \sum_{k=0}^{n-1} \Big(x+\frac{1}{n}\Big)^k $$ for 0<x<1 $\\$

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April 9th, 2019, 03:43 AM   #2
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The sum inside the limit is a geometric sequence where $\displaystyle q=x+\frac{1}{n}$ and the first term is 1 . q is the common ratio.
The limit is $\displaystyle \lim_{n\rightarrow \infty } \frac{1-q^n }{1-q}=\frac{1}{1-x}$.
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Last edited by idontknow; April 9th, 2019 at 03:46 AM.
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April 9th, 2019, 04:52 AM   #3
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Quote:
Originally Posted by idontknow View Post
The sum inside the limit is a geometric sequence where $\displaystyle q=x+\frac{1}{n}$ and the first term is 1 . q is the common ratio.
The limit is $\displaystyle \lim_{n\rightarrow \infty } \frac{1-q^n }{1-q}=\frac{1}{1-x}$.
This is not geometric. The "ratio" isn't allowed to change for each $n$.
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April 9th, 2019, 05:37 AM   #4
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For each sum, $n$ and $x$ are constants, so the sum's value is $\displaystyle \frac{1-q^n}{1 - q}$, where $q = x + \frac{\large1}{\large n}$, as already posted.

For $-1 < x < 1$, the limit exists and is $\displaystyle \frac{1}{1 - x}$.
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April 9th, 2019, 05:55 PM   #5
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Originally Posted by skipjack View Post
For each sum, $n$ and $x$ are constants, so the sum's value is $\displaystyle \frac{1-q^n}{1 - q}$, where $q = x + \frac{\large1}{\large n}$, as already posted.

For $-1 < x < 1$, the limit exists and is $\displaystyle \frac{1}{1 - x}$.
As I mentioned before, this reasoning is invalid. You are using the fact that you already know the answer to justify treating it like a geometric series. It isn't a geometric series and showing that it converges to $\frac{1}{1-x}$ is much more subtle than this.
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April 10th, 2019, 05:26 AM   #6
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The ratio can change for each n. Simply, general term is $\displaystyle a_k =(x+\frac{1}{n} )^{k-1} \; $ for $\displaystyle 0\leq k \leq n-1$.
The ratio is $\displaystyle q=\frac{a_{k+1} }{a_k }=x+\frac{1}{n} $.

Last edited by skipjack; April 10th, 2019 at 11:20 AM.
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April 10th, 2019, 11:22 AM   #7
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For each sum, it's $k$ that changes, and $n$ is the number of terms (the number of values of k), which is a constant (for that sum). Hence the usual GP formula applies. As $q$ is a function of both $x$ and $n$, evaluating $\displaystyle \lim_{n\to\infty} \frac{1-q^n }{1-q}$ shouldn't be treated as though the result is obvious.
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April 10th, 2019, 05:11 PM   #8
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Originally Posted by skipjack View Post
For each sum, it's $k$ that changes, and $n$ is the number of terms (the number of values of k), which is a constant (for that sum). Hence the usual GP formula applies. As $q$ is a function of both $x$ and $n$, evaluating $\displaystyle \lim_{n\to\infty} \frac{1-q^n }{1-q}$ shouldn't be treated as though the result is obvious.
This is just wrong. There is no other way to say it. If this were true then the following would be a geometric series
\[ \lim_{n \to \infty} \sum_{k = 0}^{n-1} n^k \]
since for fixed $n$ it has $n$ as a common ratio. But it isn't geometric and it clearly doesn't converge. I'm not sure what else to say other than to repeat myself. The question is more subtle than a simple application of a geometric series formula.
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April 10th, 2019, 06:42 PM   #9
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Since we're dealing with the limit $\lim_{n \to \infty}$ and $x \in (0,1)$, $\lim_{n \to \infty}x+\frac{1}{n} = x \in (0,1)$ So if we let $r = x+\frac{1}{n}$

$\displaystyle \lim_{n\to\infty} \sum_{k=0}^{n-1} r^k$

Since $r \in (0,1)$ , $|r| < 1$.

So I'm confused, why wouldn't this fit the bill for a geometric series?

Last edited by Greens; April 10th, 2019 at 06:47 PM. Reason: typo
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April 10th, 2019, 06:59 PM   #10
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Without the limit notation it looks like a geometric series to me, with first term $n$ and a common ratio of $n$, while computing the number of terms is not particularly straightforward. Whether the limit makes it any different is something I'd need to ponder for a longer period of time.

Very interesting thread by the way.

Last edited by greg1313; April 11th, 2019 at 03:59 AM.
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