sum from n=4 to infinity of 8/(n^2-1)

1) Let's calculate:

\sum_{n = 4}^{+∞} \frac{8}{n^{2} - 1}

⇥1.1) Let's factor in

ℝ

n^{2} - 1

⇥Factors identified:

(n + 1) \cdot (n - 1)

⇥1.2) You can decompose into partial fractions:

⇥

\frac{8}{(n + 1) \cdot (n - 1)} = \frac{A}{n + 1} + \frac{B}{n - 1}

⇥For the simple factor:

n + 1

and its root:

-1

, compute A by multiplying by the denominator:

n^{2} - 1

and evaluating at the root (the other terms cancel out):

⇥1.3) Let's solve in

ℝ

(- 1 - 1) \cdot A = 8

⇥

- 2 A = 8

⇥

A = \frac{8}{-2}

⇥

A = - \frac{8}{2}

⇥Solution found:

A = -4

⇥For the simple factor:

n - 1

and its root:

1

, compute B by multiplying by the denominator:

n^{2} - 1

and evaluating at the root (the other terms cancel out):

⇥1.4) Let's solve in

ℝ

(1 + 1) \cdot B = 8

⇥

2 B = 8

⇥

B = \frac{8}{2}

⇥Solution found:

B = 4

⇥1.5) Finally, you can substitute the values of the unknowns into the initial partial fraction decomposition:

The decomposition gives:

\frac{8}{n^{2} - 1} = \frac{4}{n - 1} - \frac{4}{n + 1}

2) Split the finite sum through

N

\sum_{n = 4}^{N} \frac{8}{n^{2} - 1} = \sum_{n = 4}^{N} \frac{4}{n - 1} - \sum_{n = 4}^{N} \frac{4}{n + 1}

Set

k = n - 1

for denominator

n - 1

and

k = n + 1

for denominator

n + 1

\sum_{k = 3}^{N - 1} \frac{4}{k} - \sum_{k = 5}^{N + 1} \frac{4}{k}

For

N >= 6

, isolate the terms at each end:

(\frac{4}{3} + \frac{4}{4} + \sum_{k = 5}^{N - 1} \frac{4}{k}) - (\sum_{k = 5}^{N - 1} \frac{4}{k} + \frac{4}{N} + \frac{4}{N + 1})

The two sums from

5

through

N - 1

are identical and cancel:

\frac{4}{3}

+ \frac{4}{4}

- \frac{4}{N}

- \frac{4}{N + 1}

3) Now take the limit as

N \to +∞

⇥3.1) Firstly, let's reduce:

\frac{4}{3}

+ \frac{4}{4}

- \frac{4}{N}

- \frac{4}{N + 1}

⇥

\frac{4 \cdot 4 + 4 \cdot 3}{3 \cdot 4} - \frac{4}{N} - \frac{4}{N + 1}

⇥ℹ

4 \cdot 4

16

‖

4 \cdot 3

12

‖

3 \cdot 4

12

⇥

\frac{16 + 12}{12} - \frac{4}{N} - \frac{4}{N + 1}

⇥

\frac{28}{12} - \frac{4}{N} - \frac{4}{N + 1}

⇥3.2) Let's evaluate the limit of

\frac{28}{12} - \frac{4}{N} - \frac{4}{N + 1}

when

N \to +∞

⇥The limit of

- \frac{4}{N}

0

⁻ when

N \to +∞

⇥ The limit of

N + 1

+∞

when

N \to +∞

⇥The limit of

- \frac{4}{N + 1}

0

⁻ when

N \to +∞

⇥So the limit of

\frac{28}{12} - \frac{4}{N} - \frac{4}{N + 1}

\frac{28}{12}

when

N \to +∞

⇥3.3) Let's evaluate:

\frac{28}{12}

▶Result:

\frac{7}{3}

4) The numeric value of:

\frac{7}{3}

is:

▷

\approx 2.3333