How this sum can be transformed into a relation with harmonic series?

Question

I am reading and textbook and it has this line :

I can not see this holds . I only get this by substituting i with (i-1):
$$\sum_{i=0}^{n-1}\frac{n}{n-i} = \sum_{i=1}^n\frac{n}{n-(i-1)}$$ Can you help me?

Answer 1

The substitution you want is $j = n - i $. That means that $i = n - j$.

The limits change as follows:

\begin{align*} i &= 0 &\Rightarrow j = n\\ i &= n-1 &\Rightarrow j = 1 \end{align*}

The sum therefore becomes

$$ \sum_{i=0}^{n-1} \frac{n}{n-i} = \sum_{j=n}^1\frac{n}{n-(n-j)} = \sum_{j=1}^n\frac{n}{j} = nH_n, $$

as desired

Answer 2

To make it clear for you, write it in inequalities. You have $i$ runs between $0$ and $n-1$ or

$$0\le i\le n-1$$

when you replace $i$ by $i-1$,

$$0\le i-1\le n-1$$

or $$ 1\le i\le n$$

you will have $i$ run between $1$ and $n$.

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Using inequalities, you can also show that

$$\sum_{i=m}^n\frac{1}{i}=\sum_{i=m}^{n}\frac{1}{n-i+m}$$

upon replacing $i$ by $n-i+m$.

or in general

$$\sum_{i=m}^n a_i=\sum_{i=m}^{n} a_{n-i+m}$$