How do you calculate the generic formula for $ \sum\limits_{i=0}^{k} \frac 1 {(n-i)}$ For the given series, how do I calculate the value of 

$$ \sum_{i=0}^{k} \frac 1 {(n-i)}$$ 

Thanks.
 A: You can write
$\begin{array}\\
\sum_{i=0}^{k} \frac 1 {(n-i)}
&=\sum_{i=n-k}^{n} \frac 1 {i}\\
&=\sum_{i=1}^{n} \frac 1 {i}-\sum_{i=1}^{n-k-1} \frac 1 {i}\\
&=H_n-H_{n-k-1}\\
\end{array}
$
where
$H_n
=\sum_{i=1}^{n} \frac 1 {i}
$.
This is the best you can do
for general $n$ and $k$.
To compute a value
where $n$
is large compared with $k$,
look up 
"Harmonic Series"
(for example, here:
https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)).
You will find that
$H_n
=\ln n +\gamma + c_n
$
where
$\gamma \approx 0.57721$
is the Euler–Mascheroni constant
and
$c_n \approx \frac1{2n}$.
The result is then,
for $n$ large compared with $k$,
$\begin{array}\\
H_n-H_{n-k-1}
&\approx \ln(n)-\ln(n-k-1)\\
&= -(-\ln(n)+\ln(n-k-1))\\
&= -(\ln(\frac{n-k-1}{n}))\\
&= -\ln(1-\frac{k+1}{n})\\
&\approx \frac{k+1}{n}\\
\end{array}
$
since
$\ln(1-x)
\approx - x
$
for small $x$.
Note that
$c_{n-k-1}-c_n
\approx \dfrac1{2(n-k-1)}-\dfrac1{2n}
=\dfrac{n-(n-k-1)}{2n(n-k-1)}
=\dfrac{k+1}{2n(n-k-1)}
$
and this is of order
$\dfrac1{n^2}$
when $k$ is small compared to $n$.
