# What does $\sum\limits_{n=1}^{N-1} \frac{1}{n} - \sum_{n=3}^{N+1} \frac{1}{n}$ simplify to?

A solution to one of the exercises in my text states:

$$\sum\limits_{n=1}^{N-1} \frac{1}{n} - \sum_{n=3}^{N+1} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} - \frac{1}{N} - \frac{1}{N+1}$$

I have no idea how to get the right hand side of the above equation.

I do realize that the left hand side involves finite sums of harmonic series which as far as I know, there is no closed form solution. So I am really puzzled as to the manipulations to be done to arrive at the right hand side. Please tell me how to get the left right hand side.

Judging by how my solution states it, it seems like this is a standard result. So please also let me know what are the properties of summation that are being used so that I can solve similar questions myself in the future.

Note that $$\sum_{n=1}^{N-1}\frac 1n=\frac 11+\frac 12+\color{red}{\frac{1}{3}+\frac14+\cdots+\frac{1}{N-1}}$$ and that $$\sum_{n=3}^{N+1}\frac 1n=\color{red}{\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{N-1}}+\frac{1}{N}+\frac{1}{N+1}$$

• I wanted to write an answer, but this just sums up too perfectly what I wanted to say. – 5xum May 4 '15 at 11:24
• Thank you so much for this answer. It is so obvious now. Could you explain how you saw this relationship? Looking at the left hand side, it was not obvious to me that there are terms going to be cancelled. – mauna May 4 '15 at 11:32
• @mauna: Well, since $1$ is near to $3$ and $N-1$ is near to $N+1$, so you can see that many terms can be cancelled. – mathlove May 4 '15 at 11:37

$\sum_{n=1}^{N-1}\frac{1}{n}-\sum_{n=3}^{N+1}\frac{1}{n}$

$=\frac{1}{1}+\frac{1}{2}+\sum_{n=3}^{N-1}\frac{1}{n}-\sum_{n=3}^{N-1}\frac{1}{n}-\frac{1}{N}-\frac{1}{N+1}$

$=\frac{1}{1}+\frac{1}{2}-\frac{1}{N}-\frac{1}{N+1}$

As for the harmonic series, this isn't a harmonic series per se, rather it is a difference of finite harmonic series with just different boundaries of summation, which should clear up why such a closed form is possible.

First thing to notice:

$$\sum_{n=1}^{N-1}\frac 1n=\frac 11+\frac 12+\cdots+\frac{1}{N-1}$$ and that $$\sum_{n=3}^{N+1}\frac 1n=\frac{1}{3}+\cdots+\frac{1}{N}+\frac{1}{N+1}$$

So what we get:

$$\left(\sum_{n=1}^{N-1}\frac 1n\right)-\left(\sum_{n=3}^{N+1}\frac 1n\right)=\frac{3}{2}-\frac{2N+1}{N^2+N}$$