you have sequence that is monotonically decreasing and bounded below by 0, hence converges to 0.
You got me. Mea culpa.hamster143 said:that's not enough. For example, 1+1/n is monotonically decreasing and it's bounded below by 0, but it does not converge to 0.
The sequence n/2^n is a mathematical sequence where each term is obtained by dividing n (a positive integer) by 2^n (2 raised to the power of n). It can be written as: 1/2, 2/4, 3/8, 4/16, 5/32, and so on.
A sequence converges when its terms get closer and closer to a single number as you go further down the sequence. In other words, the difference between each term and the limit (the number the sequence is approaching) becomes smaller and smaller.
To prove that a sequence converges, you need to show that the terms of the sequence get closer and closer to a specific number (the limit) as you go further down the sequence. This can be done through various mathematical methods such as the squeeze theorem, the ratio test, or the limit comparison test.
Proving that a sequence converges is important because it allows us to make accurate predictions about the behavior of the sequence in the long run. It also helps us understand the underlying patterns and relationships within the sequence, which can be useful in various fields of science and mathematics.
To prove that the sequence n/2^n converges to 0, we can use the limit comparison test. By comparing it to the sequence 1/2^n, which is known to converge to 0, we can show that the two sequences have the same behavior and thus, n/2^n also converges to 0. This can also be proven using the ratio test, where we can show that the limit of n/2^n as n approaches infinity is equal to 0.