sum(8/(k^2-1),k=2,+inf) — step-by-step solution

1) Let's calculate:

\sum_{k = 2}^{+∞} \frac{8}{k^{2} - 1}

⇥1.1) Let's factorize in

ℝ

k^{2} - 1

⇥For

k^{2} - 1

, we recognize a difference of two squares:

k^{2} - 1^{2}

. Apply

a^{2} - b^{2} = (a - b) \cdot (a + b)

, with

a = k

and

b = 1

⇥Factors identified:

(k + 1) \cdot (k - 1)

⇥1.2) You can decompose into partial fractions:

⇥

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

⇥For the simple factor:

k + 1

and its root:

-1

, the unknown A can be easily computed by multiplying by the denominator:

k^{2} - 1

and evaluating at the root because the other terms cancel out:

⇥1.3) Let's solve in

ℝ

(- 1 - 1) \cdot A = 8

⇥ⓘ Parentheses are computed first:

- 1 - 1

-2

⇥

- 2 A = 8

⇥ⓘ Both sides are divided by

-2

⇥

A = \frac{8}{-2}

⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{8}{-2}

- \frac{8}{2}

⇥

A = - \frac{8}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{8}{2}

-4

⇥Solution obtained:

A = -4

⇥For the simple factor:

k - 1

and its root:

1

, compute B by multiplying by the denominator:

k^{2} - 1

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

⇥1.4) Let's solve in

ℝ

(1 + 1) \cdot B = 8

⇥ⓘ Parentheses are computed first:

1 + 1

2

⇥

2 B = 8

⇥ⓘ Both sides are divided by

2

⇥

B = \frac{8}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{8}{2}

4

⇥Solution found:

B = 4

⇥1.5) Finally, substitute the unknowns back into the initial partial fraction decomposition:

⇥

\frac{8}{(k + 1) \cdot (k - 1)} = - \frac{4}{k + 1} + \frac{4}{k - 1}

The decomposition gives:

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

2) Split the finite sum through

N

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

Set

K = k - 1

for denominator

k - 1

and

K = k + 1

for denominator

k + 1

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

For

N \geq 4

, isolate the terms at each end:

(4 + \frac{4}{2} + \sum_{K = 3}^{N - 1} \frac{4}{K}) - (\sum_{K = 3}^{N - 1} \frac{4}{K} + \frac{4}{N} + \frac{4}{N + 1})

The two sums from

3

through

N - 1

are identical and cancel:

4

+ \frac{4}{2}

- \frac{4}{N}

- \frac{4}{N + 1}

3) Now take the limit as

N \to +∞

Let's evaluate the limit of

4

+ \frac{4}{2}

- \frac{4}{N}

- \frac{4}{N + 1}

when

N \to +∞

ⓘ The limit of a sum of finite limits is the sum of the limits■A number divided by a quantity that grows without bound tends to 0

The limit of

- \frac{4}{N}

0

⁻ when

N \to +∞

ⓘ The limit of a sum of finite limits is the sum of the limits■A quantity that grows without bound plus a finite or bounded quantity still grows without bound

The limit of

N + 1

+∞

when

N \to +∞

ⓘ A number divided by a quantity that grows without bound tends to 0

The limit of

- \frac{4}{N + 1}

0

⁻ when

N \to +∞

ⓘ The limit of a sum of finite limits is the sum of the limits

So the limit of

4

+ \frac{4}{2}

- \frac{4}{N}

- \frac{4}{N + 1}

4 + \frac{4}{2}

when

N \to +∞

Let's evaluate:

4 + \frac{4}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{4}{2}

2

4 + 2

ⓘ Start from

4

and move forward

2

on the number line:

4 + 2

6

ⅰ

\sum_{k = 2}^{+∞} \frac{8}{k^{2} - 1}

6

▶Result:

6