Use the ratio test: Let an = n!/en
lim n→∞ |an+1/an|
= lim n→∞ |[(n + 1)!/en+1]/(n!/en)
= lim n→∞ |[(n + 1)!/en+1](en/n!)
= lim n→∞ |[(n + 1)n!en]/(enn!e)
= (1/e) lim n→∞ (n + 1) = ∞, so the given series diverges.
Diverge!
No. ∑(1/n) diverges. It is the special infinite series known as the "harmonic series."
When you take the integral using the series as integrand, it converges if the integral worked out to be a number. If it's infinte, the series diverge.
Yes, Consider Un = (-1)^n*n = -1, 2, -3, 4, ...
The comparison test states that if a series of positive numbers converges, and in another series, each of the corresponding terms is smaller, then it too must converge. Similarly, if a series of positive numbers diverges to infinity, and another series has each of its terms greater than the corresponding terms of the other, then it too diverges.
Diverge!
No. If x tends to infinite, 1/x tends to zero.
Factorial 25 (25!) is equal to 1.5511210043 × 1025 what is 1025
No. ∑(1/n) diverges. It is the special infinite series known as the "harmonic series."
When you take the integral using the series as integrand, it converges if the integral worked out to be a number. If it's infinte, the series diverge.
Yes, Consider Un = (-1)^n*n = -1, 2, -3, 4, ...
Pi is the ratio of the circumference of a circle, to its diameter. In ancient times, it was approximated by inscribed and circumscribed polygons - the more sides the polygon had, the more accurate the approximation would be. Nowadays, infinite series are know that let you calculate pi. One well-known series of this type is:pi / 4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11... This series doesn't converge quickly; that means that you need to calculate lots of terms to get a certain degree of precision. Fortunately, other series are known that converge more quickly.
Compare a series to a known series. So take the harmonic series {1/1 + 1/2 + 1/3 + ... + 1/n}, which diverges.For each number n [n>1], LN(n) < n, so 1/(LN(n)) > 1/n. So since each term in 1/LN(n) is greater than each term in the divergent series {1/n}, then the series 1/LN(n) diverges.
tanx = 2*(sin2x - sin4x + sin6x - ... )However, be warned that this series is very slow to converge.
In calculus, you say that a series or integral converges if it has a finite value. If it does not converge, the series or integral usually diverges to infinity (that is, it does not have a finite value such as 3, -8, 67 etc.,)
Her paintings are not divided into series.
No.