Prove that the serie $\sum_{n=1}^\infty \frac{1}{n(n + a)}$ converges I was trying to solve this problem but i got stucked. I use the Radio Test to compute the convergence interval, but it doesnt works in this case, i need help...

*

*Prove that the serie $\sum_{n=1}^\infty \frac{1}{n(n + a)}$ converges

*Calculate the sum of the serie

 A: hint
Let $$u_n=\frac{1}{n(n+a)}$$
and
$$v_n=\frac{1}{n^2}$$
As you know
$$\lim_{n\to+\infty}\frac{u_n}{v_n}=1$$
$$\implies \frac{u_n}{v_n}\le 2 \text{ for great enough }  n$$
$$\implies 0< u_n\le 2v_n$$
but $$\sum v_n \text{ converges}$$
thus
$$\sum u_n \text{ converges}$$
For the sum, observe that
$$au_n = \frac 1n - \frac{1}{n+a}$$
A: If $a$ is a positive integer, it is possible to calculate the sum of the series.
Note that by partial fraction decomposition
$$
\frac1{n(n+a)} = \frac1a\left( \frac1n - \frac1{n+a} \right).
$$
So your sum becomes
$$
\frac1a\sum_{n=1}^\infty \left( \frac1n - \frac1{n+a} \right).
$$
Now, if $a$ is a positive integer, we will have a telescoping series. Let's take $a=2$ for example:
$$
\frac12\sum_{n=1}^\infty \left( \frac1n - \frac1{n+2} \right)
$$
$$
\frac12\left[\left(\frac11-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\cdots\right]
$$
Notice how in the first term we subtract $1/3$, but in the third term we add $1/3$. Those $1/3$'s will cancel out. Likewise, the $-1/4$ in the second term is cancelled by the $1/4$ in the fourth term. In fact, we can see that every term after the first two are completely canceled out, as the positive fraction is canceled with the negative fraction of the term 2 before it, and the negative fraction is canceled with the positive fraction of the term 2 after it. Only the first two positive fractions will remain.
So, for $a=2$, the sum is equal to
$$
\begin{align}
&\frac12\left(\frac11+\frac12\right)\\
=&\frac34.
\end{align}
$$
We can repeat this process for any positive integer $a$, and we will find that the sum is equal to
$$
\begin{align}
&\frac1a\left(\frac11+\frac12+\frac13+\cdots+\frac1a\right)\\
=&\frac1a\sum_{n=1}^a\frac1n.
\end{align}
$$
So, it turns out that if $a$ is an integer, we can turn the infinite sum into a finite sum, and easily compute the result.
A: Let us assume that $a$ could be any number.
$$\frac{1}{n(n + a)}=\frac 1 a\left(\frac{1}{n}-\frac{1}{n + a} \right)$$
$$S_p=\sum_{n=1}^p\frac{1}{n(n + a)}=\frac 1 a\left(H_p+H_a-H_{a+p}\right)$$ Using the asymptotics of harmonic numbers, then
$$S_p=\frac{H_a}{a}-\frac{1}{p}+\frac{a+1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$ So $S_p$ converges as long as $\frac{H_a}{a}$ exists.
This excludes all negative integer values of $a$. For any other case, rational, irrational, complex value of $a$, we whall have
$$\lim_{p\to \infty} \, S_p=\frac{H_a}{a}$$
A: Use the comparison test: your series is at least $0$ but at most$$1+\sum_{n=2}^\infty\frac{1}{n(n-1)}=1+\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac1n\right)=2,$$by telescoping.
A: Compare to integral
$$
\int_{1}^{n+1}\frac{dx}{x^2} 
$$
