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I have the following question:

Let $f:[1,2]\to\mathbb R$ defined by $f(x)=1/x$. Prove $∫_1^2 \frac{1}{x} dx=\log⁡ 2$.

In the sample answers i have given the dissection $D_n= [1,r,r^2,...,r^n] $ with $r=2^{1/n}$.

To get more experince i am trying to solve the question with a different dissection: $$D_n=[{1<\frac{n+1}{n}<\frac{n+2}{n}<\dots<\frac{2n-1}{n}<2}]$$

Using the normal method i have reached a point where i need to express $\frac{1}{n}+\frac{1}{1+n}+\dots+\frac{1}{2n-1}$ as a sum.

I have attached my working below:

Blockquote

user26857
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leahforbes
  • 111

3 Answers3

1

Just to be clear, $\frac{1}{n}+\frac{1}{1+n}+...+\frac{1}{2n-1}$ is already expressed as a sum. What you want is express it with the $\Sigma$ notation. We have $$\frac{1}{n}+\frac{1}{1+n}+\cdots +\frac{1}{n-1+n} = \sum_{k=0}^{n-1} \frac{1}{k+n}$$

jjagmath
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1

I guess you mean to express it in a compact way, since it is clearly a sum. The compact notation is$$\sum \limits _{k=0}^{n-1}\frac{1}{n+k}.$$

0

Yes, $\dfrac{1}{n}+\dfrac{1}{1+n}+\cdots +\dfrac{1}{2n-1} = \displaystyle \sum_{i=0}^{n-1} \dfrac{1}{i+n}$

Wang YeFei
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