1

$$\sum_{k=n+1}^{2n}\frac{1}{k}$$
We would like to approximate this sum using Riemann integral.
My approach is following:
Let $f:[n+1, 2n]\to \mathbb{R}$, $f(n)=\frac1n$. Let $P=(x_0, x_1,..., x_n)=(n+1, n+2, ..., 2n)$
Then $$\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{i=1}^{n}f(x_{i+1})(x_{i+1}-x_i)=\sum_{i=1}^{n}f(x_{i+1})$$
So, we have that $$\sum_{k=n+1}^{2n}\frac{1}{k}\to \int_{[n+1,2n]}\frac1xdx$$

I am not sure if I am ok, help me please understand these issues.

  • 1
    Basically you want $\int_1^2\frac1x,dx$ with the interval $[1,2]$ split into $n$ equal length subintervals. For a related discussion of these sums see for example here. – Jyrki Lahtonen Aug 28 '16 at 11:51
  • 1
    The RHS of the limit you wrote at the end of your question depends on $n$. This is a sure sign that (and in this case it happens to be due to the main reason why) something is not going well. – Did Aug 28 '16 at 18:27
  • Why interval $[1,2]$ ? Is it important ? –  Aug 29 '16 at 15:30

1 Answers1

2

Hint

$$\sum_{k=n+1}^{2n}\frac{1}{k}=\sum_{k=1}^{n}\frac{1}{k+n}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}$$

Surb
  • 55,662