Find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$

Question

find the value of $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}$ $(a>0)$

I just can analyse $\sum _{n=1}^{\infty }\:\frac{a}{n\left(n+a\right)}=a\left(\frac{1}{1}-\frac{1}{1+a}+\frac{1}{2}-\frac{1}{2+a}+\frac{1}{3}-\frac{1}{3+a}...+\frac{1}{n}-\frac{1}{n+a}\right)$

Can anyone help me? Thanks

Answer 1

It is $\psi (a + 1) + \gamma$, where $\psi$ is the logarithmic derivative of the gamma function and $\gamma$ is the Euler-Mascheroni constant, cf. $(5.7.6)$ and $(5.5.2)$ Using this fact, it follows for example that $$ \log a + \gamma + \frac{1}{{2a}} - \frac{1}{{12a^2 }} < \sum\limits_{n = 1}^\infty {\frac{a}{{n(n + a)}}} < \log a + \gamma + \frac{1}{{2a}} $$ for all $a>0$ (see $\S5.11\text{(ii)}$). Also, for $-1<a<1$, it holds that $$ \sum\limits_{n = 1}^\infty {\frac{a}{{n(n + a)}}} = \sum\limits_{k = 2}^\infty {( - 1)^k \zeta (k)a^{k - 1} } , $$ where $\zeta$ denotes Riemann's zeta function (see $(5.7.4)$).

Answer 2

Let's call the original sum $\lim_{n \to \infty} V_n$.

Asymptotic solution for $V_n$ with $a>0$: the first sum is Harmonic, so it is $\log n + O(1)$. The second sum is $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k+a} = \lim_{n \to \infty} S_n $$ Each value in the (argument, value) tuple in this sum, $(1, \frac{1}{1+a}), (2, \frac{1}{2+a}) \ldots (n, \frac{1}{n+a})$ is in fact an area of a rectangle: $r_1 = (2-1) \times \frac{1}{1+a}, r_2= (3-2) \times \frac{1}{2+a} , \ldots r_n = (n+1-n) \times \frac{1}{n+a}$, so the sum $S_n$ is equa; to the sum of areas of these rectangles.

Next step is to compare each $r_j$ to the function $f(x) = \frac{1}{x+a}$. For each interval $[1,2], (2,3], \ldots (n, n+1)$ area of $r_j$ upper-bounds integral of $f(x)$:

$$ r_j > \int_{j}^{j+1} f(x)dx = \log \frac{j+1+a}{j+a} $$ If we sum LHS and RHS of this inequality, we get the lower-bound on S_n:

$$ S_n > \sum_{j=1}^{n} > \sum_{j=1}^{n} \log \frac{j+1+a}{j+a} = \log (n+a+1) - \log (a+1) $$

As a result, you get an upper bound on the original sum:

$$ V_n < H_n - \log (n+a+1) + \log (a+1) = \log (a+1) + \gamma + \log (\frac{n}{n+a+1}) = \log (a+1) + \gamma + O(\frac{1}{n}) $$

EDIT: got the sign wrong the first time. Also $\log \frac{n}{n+a+1} = -\log (1+\frac{a+1}{n}) \sim - \frac{a+1}{n} = O(\frac{1}{n})$

Answer 3

Disclaimer: This is the same as an old answer I put for this question but it had a downvote with no explanation, so I am posting again.

This is an analytic continuation of $H_a$, which is the $a$th harmonic number. To prove it, rewrite your summand as $$\frac a{k(a+k)}=\frac 1k-\frac 1{k+a}$$And notice that the sum $$\sum_{k>0}\left(\frac 1k-\frac 1{k+a}\right)$$ starts to telescope for integer $a$.

This answer is the same as @Gary's answer, but the solutions are different. Note that $H_a=\psi(a+1)+\gamma$ where $\gamma$ is the Euler-Mascheroni Constant.

As a bit of analysis, for this definition of $H_a$ to be defined, it is required that $a\in\mathbb{R}/\mathbb{Z}^-$ we could actually use this to prove that $\Gamma(x+1)$ (which is an analytic continuation of $x!$) is also undefined at the negative integers. I could add an explanation if you want to see it.