①

Show that it is true for n = 1 first.

a₁ = 2

a₁₊₁ = a₂
= ¹⁄₅ (a²₁ + 1)
= ¹⁄₅ (2² + 1)
= ¹⁄₅ (5)
= 1 < 2

So a₂ < a₁

② Assume/suppose that it is true for n = k

i.e aₖ₊₁ < aₖ

③ Show that it is true for n = aₖ₊₁

So,

aₖ₊₂ = ¹⁄₅ (a²ₖ₊₂ + 1)


Then,

aₖ₊₂ - aₖ₊₁

= ¹⁄₅ (a²ₖ₊₁ + 1) - ¹⁄₅ (a²ₖ + 1)

= ¹⁄₅ (a²ₖ₊₁ + 1 - (a²ₖ + 1))

= ¹⁄₅ (a²ₖ₊₁ + 1 - a²ₖ - 1)

= ¹⁄₅ (a²ₖ₊₁ - a²ₖ) < 0


(Since using the assumption aₖ₊₁ < aₖ implies a²ₖ₊₁ < a²ₖ for all positive aₙ.

Eg. 4 < 5 means that 4² < 5² →16 < 25

And, why is aₙ always positive?

Because,

a₁ = 2 > 0 and aₙ₊₁ = ¹⁄₅ (a²ₙ + 1)

Since a₁ > 0 , it implies :

a²₁ > 0 (square of a positive value is still positive)

a²₁ + 1 > 0 (adding a positive value 1 to a positive value a₁² still gives a positive value)

¹⁄₅(a²₁ + 1) > 0 (multiplying by ¹⁄₅, which is positive, still results in a positive value)

So a₂ = ¹⁄₅ (a²₁ + 1) > 0 . We already found that it equals 1 above.

Then, it follows that a3 and beyond will be positive as well since the same order of operations are carried out. )


Since aₖ₊₂ - aₖ₊₁ < 0,

Then aₖ₊₂ < aₖ₊₁ . We are done with the proof.

Alternatively, when you reach this part :

= ¹⁄₅ (a²ₖ₊₁ - a²ₖ)

Factorise it using the property a² - b² = (a + b)(a - b)


= ¹⁄₅ (aₖ₊₁ + aₖ)(aₖ₊₁ - aₖ)


Since both aₖ₊₁ and aₖ are positive , (as explained earlier) then aₖ₊₁ + aₖ > 0

Combined with the assumption aₖ₊₁ - aₖ < 0 (since aₖ₊₁ < aₖ),

¹⁄₅ (aₖ₊₁ + aₖ)(aₖ₊₁ - aₖ) < 0

(Positive × positive × negative value → negative product)


So aₖ₊₂ - aₖ₊₁ < 0 and therefore aₖ₊₂ < aₖ₊₁

Or, for part ③,


aₖ₊₂ = ¹⁄₅ (a²ₖ₊₁ + 1)
aₖ₊₁ = ¹⁄₅ (a²ₖ + 1)


Since aₖ₊₁ < aₖ and both are positive,

a²ₖ₊₁ < a²ₖ

a²ₖ₊₁ + 1 < a²ₖ + 1

¹⁄₅ (a²ₖ₊₁ + 1) < ¹⁄₅ (a²ₖ + 1)

aₖ₊₂ < aₖ₊₁

(Shown)