# Computing a sum of an infinite series via partial decomposition

Consider the sequence defined by $u_0=0$ and $u_n=\frac{4n+1}{(2n+3)(4n^2-1)}$. The exercise asks to show that the series $\sum_{n\geq 0}{u_n}$ is convergent, which is clear since $u_n\sim_\infty \frac{4n}{8n^3}=\frac{1}{2n^2}$. At this stage i have two questions:

1) Why did we single out $u_0$ from the definition of $u_n$ which is well defined for $n=0$

2) If someone is not aware of the notion of equivalence, what is the procedure to find a convergent $p$-series to which we will compare $u_n$.

The rest of the exercice asks the partial fraction decomposition of $u_n$ and i find that $u_n=\frac{a}{2n-1} + \frac{b}{2n+1} + \frac{c}{2n+3}$ with $a=3/8$, $b=1/4$, $c=-5/8$, But then he asks to deduce the sum $\sum_{n= 0}^{\infty}{u_n}$.

3) How is it possible to deduce the sum from the partial decomposition where the sum of each fraction does not exist, as the corresponding series $\sum_{n\geq 0}{\frac{a}{2n-1}}$, $\sum_{n\geq 0}{\frac{b}{2n+1}}$ and $\sum_{n\geq 0}{\frac{c}{2n+3}}$ diverge ??

Thank you for your help!!

• 1) Don't know. Maybe because you get something nicer for the sum with $u_0 = 0$. 2) Don't know either. 3) Partial sums, telescoping. – Daniel Fischer Oct 20 '13 at 15:37
• How does the change of $u_0$ alter the sum of the infinite series? I remember seeing a result saying that the sum does not change if we change the values of a finite number of terms of the series, i'm not sure about the exact statement!! – palio Oct 20 '13 at 15:46
• It changes the sum by a constant. If the sum with $u_0 = 0$ were, say, $\pi$, with $u_0 = \frac{4\cdot 0+1}{(2\cdot 0+3)(4\cdot 0^2-1)} = -\frac13$ it would be $\pi -\frac13$, not as nice. – Daniel Fischer Oct 20 '13 at 15:52
• I did the telescoping and the sum is $7/12$, taking $u_0=0$ was just to simplify telescoping and having less elements to put together.. Thank you Daniel for your help!! – palio Oct 20 '13 at 16:12