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Task: Using a definite integral find the value of: $$\lim_{n\rightarrow \infty }(\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n})$$

My Attempt: I began by writing out the sequence as a summation, where I afterwards isolated the $n$ sub-intervals multiplication:

$$ \lim_{n\rightarrow \infty }\sum_{i=0}^{n}{\frac{1}{n+i}} = \lim_{n\rightarrow \infty }\sum_{i=0}^{n}{\frac{n}{n+i}\frac{1}{n}} $$ Here I encountered an unfamiliar situation with $i$ in the denominator and not in the numerator. Further investigation lead me to harmonic numbers, which is something I haven't covered yet and shouldn't be required.

In attempting to solve this task I have found the following resource on the limit definition of a definite integral.

zgfico
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5 Answers5

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Hint: $$\frac{n}{n + i} = \frac{1}{1 + i/n}$$

Now apply the Riemann-integral.

Dominik
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$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

$$\lim_{n\to\infty}\sum_{r=1}^n\dfrac1{n+r}=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\dfrac1{1+\dfrac rn}$$

So, $f\left(\dfrac rn\right)=\dfrac1{1+\dfrac rn}, f(x)=?$

See also : The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

Community
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lab bhattacharjee
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Hint: $$\sum_{i = n}^{2 \, n} \frac{1}{i} \le \int_{n-1}^{2\,n} \frac1x \, \mathrm{d}x$$ (just look at the graph of $1/x$).

Similarly, you get a lower bound.

gerw
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There's a better way: rewrite the sum as $H_{2n} - H_{n-1} \sim \log 2n - \log (n-1) = \log 2 + \log \frac{n}{n-1} \to_n \log 2$

Alex
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    You would need to establish first that $H_n - \log(n)$ converges to be able to use this argument (at the level this question is asked I don't think this can be taken as common knowledge). This is imo more complicated than using the Riemann sum argument. – Winther Jan 06 '16 at 14:09
  • 'need to establish that $H_n - \log (n)$ converges' - this follows directly from the definition of the harmonic sum, as the next term is a constant. – Alex Jan 06 '16 at 14:12
  • The definition of the harmonic sum is $H_n = 1 + 1/2 + \ldots + 1/n$. I don't see how it follows directly from this. – Winther Jan 06 '16 at 14:16
  • OK sorry, I mean asymptotic expansion: $H_n \sim \log_n + \gamma + o(1)$ – Alex Jan 06 '16 at 14:27
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    I agree that this argument is as simple as it gets once you know that $H_n = \log(n) + \gamma + O(1/n)$. The only point I was trying to make was that I don’t think one can assume that this is known at the level this question is asked (i.e. what we would expect OP to know). Anyway there is nothing wrong with the argument, I just think it’s using a bit too high level tools (and therefore disagree with "there is a better way" :) – Winther Jan 06 '16 at 14:29
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Apparently I can continue this task by multiplying both the nominator and denominator by $\frac{1}{n}$. Which gives me the following: $$ \lim_{n\rightarrow \infty }\sum_{i=0}^{n}{\frac{n\frac{1}{n}}{(n+i)\frac{1}{n}}\frac{1}{n}} = \lim_{n\rightarrow \infty }\sum_{i=0}^{n}\frac{1}{1+\frac{i}{n}}\frac{1}{n} $$

Here we can take that the sampling point is $\frac{i}{n}$ which leads to this equation Riemann sum for the following function $f(x)=\dfrac{1}{1+x}$ on the range from $I=[0,1]$.

So, the limit in the equals: $$\int_0^1{\dfrac{1}{1+x} \mathrm dx}= \ln 2$$

zgfico
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