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I need to calculate $$\lim_{n \to \infty}\frac{1}{n}\left(\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n}+\cdots+\frac{3n}{4n}\right).$$ It can be written as $$\lim_{n \to \infty}\frac{1}{n^2}\left(\frac{1}{\frac{1}{n}+1} + \frac{1}{\frac{2}{n}+1} + \frac{1}{\frac{3}{n}+1}+\cdots+\frac{1}{\frac{4n}{3n}}\right).$$ Now I need to write it as $\frac{1}{n}\sum_{?}^{?}{?}$ and ultimately in definite integral form so that I can evalute the sum of this limit.

dfeuer
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Tesla
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  • I'm a little confused about the terms in the sum. For any $n \geq 1$, the largest term is the last one $3n/4n=3/4$. So, how many terms come in the sum before that? Any chance that the last term is supposed to be something of the form $\frac{1}{m n}$ where $m \in \mathbb{N}$ is some positive integer? – Tom Oct 31 '13 at 19:01
  • no it's $\frac{3n}{4n}$ i doubt we can just directly write $\frac{3n}{4n}=\frac{3}{4}$ when n is tending towards infinity or i am wrong ? – Tesla Oct 31 '13 at 19:04
  • Yes, $n$ is tending toward $\infty$. But, for each finite $n$, $\frac{3n}{4n}=\frac{3}{4}$. So, @Tom is right. – J126 Oct 31 '13 at 19:06
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    Well, certainly you can take a limit of a constant term (for example $\lim_{n\to \infty} 2 = 2$), but the real problem is that it doesn't fit the pattern of the previous terms $\frac{1}{k+n}$, so I can't interpret what the sum really is. – Tom Oct 31 '13 at 19:06
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    @Tesla The statement doesn't make too much sense from my point of view. $1/x$ is a decreasing function, so for $n\in\Bbb N$, $\frac34>\frac1{1+n}>\frac1{2+n}>\frac1{3+n}>\dots$, so there is no way that $\frac34=\frac1{k+n}$ for some $k$. – Mario Carneiro Oct 31 '13 at 19:07
  • @MarioCarneiro i don't know maybe that's how problem was written in my book so i asked it here how to do it and i m still unable to do it – Tesla Oct 31 '13 at 19:10
  • @Tesla Is the first expression exactly verbatim what your book says? If so you should ask your professor or check the errata. – Mario Carneiro Oct 31 '13 at 19:13

2 Answers2

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Hint:
$$ \frac{1}{n}\sum_{k=1}^{mn} \frac{1}{\frac{k}{n} + 1} \approx \int_0^m \frac{1}{x+1}dx $$

Tom
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Try Riemann sums. Here is a related problem. Here is how you write the sum

$$ \frac{1}{n}\sum_{k=1}^{3n}\frac{1}{k+n}. $$

Community
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Mhenni Benghorbal
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