Question

Help on math! First correct answer will get brainliest answer

Answer 1

Let's look at the terms of the sequence for different values of n, starting from 0.

a₀ = 0/1 = 0

a₁ = 1/(1 +1) = 1/2

a₂ = 2/(2 + 1) = 2/3

a₃ = 3/(3 + 1) = 3/4

a₄ = 4/5

a₅ = 5/6

a₆ = 6/7

a₇ = 7/8

a₈ = 8/9

a₉ = 9/(9+1) = 9/10

a₁₀ = 10/(10+1) = 10/11 and so on

Let's look at the terms. As n gets bigger the terms gets closer to 1. We started with 0, then to one half, then two thirds, and as we get larger and larger, the terms are getting closer to 1. Let's choose a really big n and see.

a₉₉₉ = 999/(999 + 1) = 999/1000

Or an even bigger n:

a₉₉₉₉₉₉₉₉₉ = 999999999/(999999999 + 1) = 999999999/1000000000ⁿ

So as n gets really really large - and close to infinity - the terms get closer to 1.