Problem 40
Question
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1+(-1)^{n}}{n^{2}}\)
Step-by-Step Solution
Verified Answer
The sequence \(a_{n}=\frac{1+(-1)^{n}}{n^{2}}\) converges and its limit is 0.
1Step 1: Analyze the given sequence
First, observe and analyze the given sequence \(a_{n}=\frac{1+(-1)^{n}}{n^{2}}\). Given that \((-1)^{n}\) toggles between -1 and 1 depending on whether 'n' is odd or even respectively, it changes the numerator of the sequence between 0 and 2.
2Step 2: Split into subsequences
Split the sequence into two subsequences. One, where 'n' is odd (n=2k-1, k as a natural number) and one where 'n' is even (n=2k). For 'n' odd, \((-1)^{n} = -1\) and thus \(a_{n}=0\). For 'n' even, \((-1)^{n} = 1\) and hence \(a_{n} = \frac{2}{n^{2}}\).
3Step 3: Determine the limits
Now that we understand the behavior of the sequence better, the next step is to calculate the limits as 'n' approaches infinity for each subsequence. For the subsequence where 'n' is odd, the limit is obviously 0 (because it's a constant sequence). For the subsequence where 'n' is even, we need to mathematically determine the limit: \(\lim_{{n \to \infty}}{\frac{2}{n^{2}}}=0\). The limit of the sequence is therefore 0.
4Step 4: Determine convergence or divergence
Both subsequences tend to 0 as 'n' approaches infinity, showing that the original sequence is indeed convergent, and that its limit is 0.
Key Concepts
Limits of a SequenceSubsequencesAlternating SeriesInfinite Limits
Limits of a Sequence
Understanding the limit of a sequence is essential to analyzing its behavior as the sequence extends towards infinity. In essence, if a sequence has a limit, this means that as you progress through the terms of the sequence (taking on larger and larger values of the index, typically symbolized by 'n'), the terms of the sequence get closer and closer to a specific value.
The limit of a sequence is defined mathematically as a number 'L' such that for every positive number \( \) (no matter how small), there exists a natural number 'N' where for all \( n > N \) , the distance between \( a_n \) and 'L' is less than \( \). If such a number 'L' exists, the sequence converges to 'L'. If no such 'L' can be found, the sequence is said to diverge.
In the given exercise, the limit of the sequence \( a_{n}=\frac{1+(-1)^{n}}{n^{2}} \) as \( n \) goes to infinity is sought after. The crucial step is recognizing that the sequence behaves differently depending on whether 'n' is even or odd, which highlights the importance of examining subsequences, as was done in the provided solution.
The limit of a sequence is defined mathematically as a number 'L' such that for every positive number \( \) (no matter how small), there exists a natural number 'N' where for all \( n > N \) , the distance between \( a_n \) and 'L' is less than \( \). If such a number 'L' exists, the sequence converges to 'L'. If no such 'L' can be found, the sequence is said to diverge.
In the given exercise, the limit of the sequence \( a_{n}=\frac{1+(-1)^{n}}{n^{2}} \) as \( n \) goes to infinity is sought after. The crucial step is recognizing that the sequence behaves differently depending on whether 'n' is even or odd, which highlights the importance of examining subsequences, as was done in the provided solution.
Subsequences
A subsequence is a sequence that is derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Subsequences retain their original 'order', but not necessarily their 'spacing'.
This is particularly helpful in analyzing sequences that exhibit different behavior at different stages, such as the sequence in our exercise. By splitting it into two subsequences based on whether 'n' is odd or even, it makes the analysis more tractable. For this problem, we consider subsequences for odd and even 'n' values separately.
This is particularly helpful in analyzing sequences that exhibit different behavior at different stages, such as the sequence in our exercise. By splitting it into two subsequences based on whether 'n' is odd or even, it makes the analysis more tractable. For this problem, we consider subsequences for odd and even 'n' values separately.
Determining Limits of Subsequences
For the subsequence where 'n' is odd, the terms are constant (\( a_{n}=0 \)), which trivially converges to 0. For the subsequence where 'n' is even, the terms \( a_{n} = \frac{2}{n^{2}} \) are a classic example of a null sequence, where the terms get closer and closer to 0 as 'n' increases, showing convergence to 0.Alternating Series
An alternating series is one in which the signs of the terms alternate between positive and negative. Recognizing an alternating series is key to determining convergence in many cases. The sequence at hand, \( a_{n}=\frac{1+(-1)^{n}}{n^{2}} \), involves an alternating term, \( (-1)^{n} \), which toggles between -1 and 1.
While the sequence in our exercise isn't a 'pure' alternating series, as the signs do not strictly alternate due to the presence of zero in the sequence when 'n' is odd, understanding the behavior of this alternating component is critical to solving the problem. It's pivotal in splitting the sequence into two subsequences and ultimately in establishing the convergence to 0, as both subsequences exhibit non-alternating behavior and converge to the same limit.
While the sequence in our exercise isn't a 'pure' alternating series, as the signs do not strictly alternate due to the presence of zero in the sequence when 'n' is odd, understanding the behavior of this alternating component is critical to solving the problem. It's pivotal in splitting the sequence into two subsequences and ultimately in establishing the convergence to 0, as both subsequences exhibit non-alternating behavior and converge to the same limit.
Infinite Limits
The concept of infinite limits is useful when dealing with sequences in which the terms grow without bound as 'n' tends to infinity; this indicates divergence. However, if the terms approach a number 'L', the sequence has a limit and thus converges.
In the case of \( a_{n}=\frac{1+(-1)^{n}}{n^{2}} \), the terms do not grow without bound; rather, they tend to 0. The notion of an infinite limit helps us exclude the possibility of the sequence diverging to infinity or negative infinity. Verification of the convergence to a finite number, again, embeds the necessity to engage with the concept of limits; specifically by showing that both subsequences have a finite limit, we can deduce that the entire sequence converges to 0.
In the case of \( a_{n}=\frac{1+(-1)^{n}}{n^{2}} \), the terms do not grow without bound; rather, they tend to 0. The notion of an infinite limit helps us exclude the possibility of the sequence diverging to infinity or negative infinity. Verification of the convergence to a finite number, again, embeds the necessity to engage with the concept of limits; specifically by showing that both subsequences have a finite limit, we can deduce that the entire sequence converges to 0.
Other exercises in this chapter
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