[2024/05/07] Evaluate converging sum ( lemmy.world )

zkfcfbzr , 1 month ago

solution

With partial fractions:

1/(n + n²) = 1/(n(n+1)) = A/n + B/(n+1)

A(n+1) + Bn = 1

n = 0 gives A = 1, n = -1 gives B = -1

1/(n+n²) = 1/n - 1/(n+1)

Σ (n = 1 to ∞) 1/(n+n²) = Σ (n = 1 to ∞) 1/n - Σ (n = 1 to ∞) 1/(n+1)

= Σ (n = 1 to ∞) 1/n - Σ (n = 2 to ∞) 1/n

= 1/1 + Σ (n = 2 to ∞) 1/n - Σ (n = 2 to ∞) 1/n

= 1

Guessing this is the standard solution

WhoresonWells , 1 month ago

Since this is everyone's favorite example of telescoping sums, let's do it another way just for giggles.

Combinatorial proof

The denominator is P(n+1, 2) which is the number of ways for 2 specified horses to finish 1st and second in an n+1 horse race.
So imagine you're racing against horses numbered {1, 2, 3, ....}.
Either you win, which has probability 0 in the limit, or there is a lowest numbered horse, n, that finishes ahead of you.
The probability that you beat horses {1,2, ... , n-1} but lose to n is (n-1)! / (n+1)! or P(n+1, 2) or 1/(n^2^+n), the nth term of the series.
Summing these mutually exclusive cases exhausts all outcomes except the infinitesimal possibility that you win.
Therefore the infinite sum is exactly 1.

---

siriusmart OP Mod , 1 month ago (edited 1 month ago)

Hint 1:

spoiler

expand the expression

---

Hint 2:

spoiler

partial fractions

---

Solution:

spoiler

Link: https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd%20solutions/2024-05-07_infinite-sum.html
https://lemmy.world/pictrs/image/b10bbfbe-728e-4873-ab6f-1758e61bbf13.png