# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 3/(1^2*2^2)+5/(2^2*3^2)+...+(2n+1)/(n^2(n+1)^2)+...?

Feb 4, 2017

The partial sum is $= \frac{n \left(n + 2\right)}{n + 1} ^ 2$
The series converge and the sum is $= 1$

#### Explanation:

Let's perform a partial fraction decomposition

$\frac{2 n + 1}{{n}^{2} {\left(n + 1\right)}^{2}} = \frac{A}{n} ^ 2 + \frac{B}{n} + \frac{C}{n + 1} ^ 2 + \frac{D}{n + 1}$

$= \frac{A {\left(n + 1\right)}^{2} + B n {\left(n + 1\right)}^{2} + C {n}^{2} + C {n}^{2} \left(n + 1\right)}{{n}^{2} {\left(n + 1\right)}^{2}}$

As the denominators are the same, we compare the numerators

$2 n + 1 = A {\left(n + 1\right)}^{2} + B n {\left(n + 1\right)}^{2} + C {n}^{2} + D {n}^{2} \left(n + 1\right)$

Let $n = 0$, $\implies$, $1 = A$

Let $n = - 1$, $\implies$, $- 1 = C$

Coefficients of ${n}^{3}$, $\implies$, $0 = B + D$

Coefficients of $n$, $\implies$, $2 = 2 A + B$

$B = 2 - 2 A = 2 - 2 = 0$

$D = - B = 0$

So,

$\frac{2 n + 1}{{n}^{2} {\left(n + 1\right)}^{2}} = \frac{1}{n} ^ 2 - \frac{1}{n + 1} ^ 2$

Now we can calculate the sums

$n = 1$,$\implies$,${S}_{1} = 1 - \cancel{\frac{1}{4}}$

$n = 2$,$\implies$,${S}_{2} = \cancel{\frac{1}{4}} - \cancel{\frac{1}{9}}$

$n = 3$, $\implies$, ${S}_{3} = \cancel{\frac{1}{9}} - \cancel{\frac{1}{16}}$
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$n = n - 1$, $\implies$, ${S}_{n - 1} = \cancel{\frac{1}{n - 1} ^ 2} - \cancel{\frac{1}{n} ^ 2}$

$n = n$, $\implies$, ${S}_{n} = \cancel{\frac{1}{n} ^ 2} - \frac{1}{n + 1} ^ 2$

So,

${S}_{1} + S - 2 + {S}_{2.} \ldots \ldots \ldots . {S}_{n} = 1 - \frac{1}{n + 1} ^ 2$

$= \frac{{\left(n + 1\right)}^{2} - 1}{n + 1} ^ 2$

$= \frac{n \left(n + 2\right)}{n + 1} ^ 2$

${\lim}_{n \to + \infty} \frac{n \left(n + 2\right)}{n + 1} ^ 2$

$= {\lim}_{n \to + \infty} {n}^{2} / {n}^{2} = 1$