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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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 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. Tags limit, serie, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post yu123 Calculus 1 February 4th, 2015 08:35 AM Gandalf Real Analysis 2 April 18th, 2014 02:26 AM galc127 Real Analysis 5 November 9th, 2013 10:41 AM xpoisnp Calculus 1 January 27th, 2012 12:05 PM aaron-math Calculus 3 November 22nd, 2011 08:09 AM

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