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How do I evaluate $$\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)\ ?$$

I'm stuck. I tried using the sandwich theorem but I was getting nowhere. Since $n \lt n+k \lt 2n$, I think it should be $ \infty$ .

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  • $\begingroup$ Would this, by any chance, be a problem in a calculus book, shortly after discussion of Riemann sums? $\endgroup$ – GEdgar Mar 7 '15 at 14:07
  • $\begingroup$ Can We Solve it using sandwich Theorem, If Yes Then plz explain here, Thanks. $\endgroup$ – juantheron Aug 10 '15 at 3:58
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It can be rewritten as

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

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

And here you recognize a Riemann sum. So the limit is

$$\int_0^1 \frac{1}{1+x} dx$$

Edit : when something can be written as $\frac{1}{n} \sum \cdots$, a sum of Riemann may be hidden inside.

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  • $\begingroup$ So what's the general form of a riemann sum? $\endgroup$ – N S Mar 7 '15 at 10:38
  • $\begingroup$ It's not always obvious you need to find a Riemann sum. The "general form" is $$\frac{1}{n} \sum_{k=0}^n f(\frac{k}{n})$$ But it's not always obvious that something can be written like that. As an exemple, can you find how to make appear a Riemann sum with this expression? $$ n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}}$$ $\endgroup$ – Tryss Mar 7 '15 at 11:54
  • $\begingroup$ By taking logarithm? @tryss $\endgroup$ – N S Mar 7 '15 at 12:17
  • $\begingroup$ Yes, you take the logarithm, and you make some manipulations to transform this nasty expression in a gentle Riemann sum. But would you think about Riemann sum if you stumbled onto this limit "in the wild"? $\endgroup$ – Tryss Mar 7 '15 at 12:19
  • $\begingroup$ No I wouldnt!! But thanks to you now I'd remember $\endgroup$ – N S Mar 7 '15 at 12:28
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More directly, multiplying by $1/n$ each term inside the parenthesis we get

$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=$$ $$\lim_{n\to\infty}1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}-1-\frac{1}{2}-\frac{1}{3}-\cdots-\frac{1}{n}=\\\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k},$$ which, since the harmonic series grows logarithmically, is the same as $$\lim_{n\to\infty}\log(2n)-\log n=\log 2.$$

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  • $\begingroup$ Didnt get the second step? How can you arrive to that? $\endgroup$ – N S Mar 7 '15 at 13:45
  • $\begingroup$ @NS I've added a middle step, is it clearer now? $\endgroup$ – Vincenzo Oliva Mar 7 '15 at 14:03
  • $\begingroup$ @NS Glad I could help! $\endgroup$ – Vincenzo Oliva Mar 7 '15 at 14:20
  • $\begingroup$ @NS In all calmness, may I know why you changed your decision as for the accepted answer? ^_^ Perhaps I could learn something more on how to write an answer. $\endgroup$ – Vincenzo Oliva Mar 7 '15 at 17:05
  • $\begingroup$ Yours is really truly intuitive. But the above answer I chose as its more formal. I wish I could've chosen both. :( $\endgroup$ – N S Mar 8 '15 at 12:28
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using digamma function we have $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}=\psi(n+1)-\psi(2n+1)$$ but then using the following asymptotic expansion $$\psi(x)= \ln(x) - \sum_{k=1}^\infty \frac{B_k}{k x^k}$$ with type two Bernoulli numbers $B_k$, we have \begin{align} \lim_{n \to \infty}\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}&=\lim_{n \to \infty}\Big(\psi(n+1)-\psi(2n+1)\Big)\\ &=\ln2 \end{align}

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    $\begingroup$ Nice Solution Math-fun. $\endgroup$ – juantheron Aug 10 '15 at 3:37
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Notice, we have $$\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$$

$$=\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{n}{n+r} \right)$$ $$=\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}} \right)$$

Let $\displaystyle \frac{r}{n}=x\implies \lim_{n\to \infty}\frac{1}{n}=dx\to 0$ $$ \text{lower limit of }\ x=\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{1}{n}=0$$ $$ \text{upper limit of }\ x=\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{n}{n}=1$$ Now, we have $$\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}} \right)=\int_{0}^{1}\left(\frac{1}{1+x} \right)dx$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$$$=\ln (2)-\ln(1)=\ln 2$$

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    $\begingroup$ This is quite nearly a restatement of the accepted (and top) answer. $\endgroup$ – Cameron Williams Aug 10 '15 at 3:13
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Since

$\begin{array}\\ \frac1{k} &> \int_k^{k+1} \frac{dx}{x} > \frac1{k+1}\\ \sum_{k=n+1}^{2n}\frac1{k} & > \sum_{k=n+1}^{2n}\int_k^{k+1} \frac{dx}{x} > \sum_{k=n+1}^{2n}\frac1{k+1}\\ or\\ \sum_{k=n+1}^{2n}\frac1{k} & > \int_{n+1}^{2n+1} \frac{dx}{x} > \sum_{k=n+1}^{2n}\frac1{k}-\frac1{n+1}+\frac1{2n+1}\\ \end{array} $

Since $\int_{n+1}^{2n+1} \frac{dx}{x} =\ln(2n+1)-\ln(n+1) =\ln(2-\frac1{n+1}) =\ln(2)-\ln(1-\frac1{2n+2}) $, if $d(n) =\ln(2)-\sum_{k=n+1}^{2n}\frac1{k} $, $d(n) < \ln(1-\frac1{2n+2}) < 0 $ and $d(n) >\ln(1-\frac1{2n+2})-\frac1{n+1}+\frac1{2n+1} $.

Setting $k=2n+1$ in $\frac1{k} > \int_k^{k+1} \frac{dx}{x} > \frac1{k+1} $ we get $\frac1{2n+1} > -\ln(1-\frac1{2n+1}) > \frac1{2n+2} $. Therefore $d(n) >-\frac1{2n+1}-\frac1{n+1}+\frac1{2n+1} =-\frac1{n+1} $. so $$-\frac1{n+1} < \ln(2)-\sum_{k=n+1}^{2n}\frac1{k} < 0. $$

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