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 April 9th, 2019, 01:06 AM #1 Member   Joined: Mar 2016 From: Sweden Posts: 35 Thanks: 4 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
 April 9th, 2019, 03:43 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 713 Thanks: 96 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 $\displaystyle \lim_{n\rightarrow \infty } \frac{1-q^n }{1-q}=\frac{1}{1-x}$. Thanks from topsquark and matteamanda Last edited by idontknow; April 9th, 2019 at 03:46 AM.
April 9th, 2019, 04:52 AM   #3
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Quote:
 Originally Posted by idontknow 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 $n$.

 April 9th, 2019, 05:37 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 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 $\displaystyle \frac{1}{1 - x}$. Thanks from idontknow
April 9th, 2019, 05:55 PM   #5
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 Originally Posted by skipjack 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.

 April 10th, 2019, 05:26 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 713 Thanks: 96 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.
 April 10th, 2019, 11:22 AM #7 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 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. Thanks from topsquark
April 10th, 2019, 05:11 PM   #8
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Quote:
 Originally Posted by skipjack 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.

 April 10th, 2019, 06:42 PM #9 Member     Joined: Oct 2018 From: USA Posts: 98 Thanks: 70 Math Focus: Algebraic Geometry 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
 April 10th, 2019, 06:59 PM #10 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond 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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