Table of Contents
Does the series n sin 1 n converge or diverge?
We also know that 1n diverges at infinity, so sin(1n) must also diverge at infinity.
Does the series of sin 1 n converge?
Hence the two series Sigma [sin(1/n)] and Sigma(1/n) have the same convergence behaviour by limit comparison test, for series of positive terms But we know that the harmonic series Sigma(1/n) diverges. Hence the given series Sigma[sin(1/n) also diverges.
How does the limit comparison test work?
The Limit Comparison Test
- If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
- If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
Does 1 LNN converge?
Answer: Since ln n ≤ n for n ≥ 2, we have 1/ ln n ≥ 1/n, so the series diverges by comparison with the harmonic series, ∑ 1/n.
Does sin 1 n = 1 2 n converge absolutely?
No, it does not converge absolutely. Note sin 1 n ∼ 1 n for large n. This implies sin 1 n ≥ 1 2 n ≥ 0 for large n. But we know that ∑ 1 2 n = 1 2 ∑ 1 n = + ∞ and so by comparison ∑ sin 1 n = + ∞. However, the series converges conditionally.
What is the value of sin 1 n for large N?
1 n ∼ 1 n for large n. This implies sin 1 n ≥ 1 2 n ≥ 0 for large n. But we know that ∑ 1 2 n = 1 2 ∑ 1 n = + ∞ and so by comparison ∑ sin 1 n = + ∞. However, the series converges conditionally. This is an immediate consequence of the alternating series test. | sin 1 n is positive and monotonically decreasing.
Is ∞ ∑ N = 1LN(1 + 1 n) a convergent series?
So, the series ∑ 1 n and ∑ln(1 + 1 n) are both convergent or divergent. Since N ∑ n = 1ln(1 + 1 n) = ln(N + 1) − ln(1) = ln(N + 1). Thus ∞ ∑ n = 1ln(1 + 1 n) is divergent and so is ∞ ∑ n = 11 n.
Why do series have to converge to zero to converge?
Again, as noted above, all this theorem does is give us a requirement for a series to converge. In order for a series to converge the series terms must go to zero in the limit. If the series terms do not go to zero in the limit then there is no way the series can converge since this would violate the theorem.