evaluate the expression [1]:
$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)} $$
where $x$ is a real number, $0\le x\le1$, and $x$ is rounded to 3 digits.
For example, when $x=0.500$, the expression is [2]:
$$\left(\frac11 -\frac1{1.5}\right)+\left(\frac12 -\frac1{2.5}\right)+\left(\frac13 -\frac 1{3.5}\right) + ...$$
For a given $x$, how can I evaluate it?
The answer must be rounded to more than 12 digits.
Answer
$$\sum\limits_{n = 1}^\infty {\left( {\frac{1}{n} - \frac{1}{{n + x}}} \right)}=\sum\limits_{n = 1}^\infty \frac{x}{n(n+x)}$$
It is known that the sum is
$$\sum\limits_{n = 1}^\infty \frac{x}{n(n+x)}=\psi(x+1)+\gamma$$
where $\psi$ is the digamma function
$\gamma$ is the Euler constant .
Added :
We can easily prove that for $x=0.5$ we have
$$\psi(1.5)=2-\gamma -\log(4)$$
Hence the sum is equal to
$$2-\log(4)$$
No comments:
Post a Comment