Question

Let \(a_n=\frac{1+2^{-2}+...+n^{-2}}{n}\) for n ∈ \(\N\). Then

• A. both the sequence (a_n) and the series \(\sum\limits_{n=1}^{\infin}a_n\) are convergent
• B. the sequence (a_n) is convergent but the series \(\sum\limits_{n=1}^{\infin}a_n\) are NOT convergent
• C. both the sequence (a_n) and the series \(\sum\limits_{n=1}^{\infin}a_n\) are NOT convergent
• D. the sequence (a_n) is NOT convergent but the series \(\sum\limits_{n=1}^{\infin}a_n\) is convergent

Answer 1

The Correct Option is B

To determine the convergence of the sequence \(a_n\) and the series \(\sum_{n=1}^{\infty}a_n\), we need to analyze both these mathematical entities step-by-step.

Step 1: Analyze the Sequence \(a_n\)

The sequence is defined as:

\(a_n = \frac{1 + 2^{-2} + \cdots + n^{-2}}{n}\)

This sequence can be rewritten using the summation notation:

\(a_n = \frac{\sum_{k=1}^{n} k^{-2}}{n}\)

As \(n\) tends to infinity, the numerator \(\sum_{k=1}^{n} k^{-2}\) approximates the sum of the series \(\sum_{k=1}^{\infty} k^{-2}\), which is known to converge to a finite number (specifically, \(\frac{\pi^2}{6}\)). Therefore, the sequence can be expressed as:

\(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^{-2}}{n} = 0\)

Thus, the sequence \(a_n\) converges to 0.

Step 2: Analyze the Series \(\sum_{n=1}^{\infty} a_n\)

Consider the series:

\(\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\sum_{k=1}^{n} k^{-2}}{n}\)

Since \(a_n = \frac{\sum_{k=1}^{n} k^{-2}}{n}\) behaves similarly to the harmonic series when considered for its tail behavior (as \(\frac{1}{n}\) remains), we can use the standard comparison with the harmonic series:

The harmonic series is \(\sum \frac{1}{n}\), which diverges. Because \(\sum_{k=1}^{n} k^{-2} \leq n\) for large \(n\), \(a_n\) behaves similarly to \(\frac{1}{n}\) but without damping that ensures convergence.

Therefore, the series \(\sum_{n=1}^{\infty} a_n\) diverges.

Conclusion

After the analysis, we conclude that:

• The sequence \(a_n\) converges to 0.
• The series \(\sum_{n=1}^{\infty} a_n\) diverges.

Thus, the correct answer is: the sequence (a_n) is convergent but the series \(\sum_{n=1}^{\infty} a_n\) is NOT convergent.