How do you calculate the generic formula for $ \sum\limits_{i=0}^{k} \frac 1 {(n-i)}$

Question

For the given series, how do I calculate the value of

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

Thanks.

Answer 1

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 $n$.