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

Question

---

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.

Answer 1

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

Now apply the Riemann-integral.

Answer 2

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

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

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

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

Answer 3

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.

Answer 4

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

Answer 5

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: $$\int_0^1{\dfrac{1}{1+x} \mathrm dx}= \ln 2$$