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The series does not converge. Let s(n) be the partial sum of the series. s(1) = 4, s(2) = 1, s(3) = 0, s(4)=4, etc. It is plain to see that s(n) is periodic, i.e. that s(n) = s(n+3). Hence, s(n) does not approach a limit as n -> infinity, so the series does not approach a fixed value. Therefore, it diverges.
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.
584 its [ (n-1)^3 + (n-1)^2 + (n-1) ]
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Yes, Consider Un = (-1)^n*n = -1, 2, -3, 4, ...
Each number in the series is being multiplied by (2 + (n-1)) where n = the term in the series 3 x 2 = 6 6 x 3 = 18 18 * 4 = 72 72 x 5 = 360
To find the sum of the series of the first 20 positive integers ending in 3, we first need to identify the pattern. The series would be 3, 13, 23, 33, ..., 193. This is an arithmetic series with a common difference of 10. To find the sum, we can use the formula for the sum of an arithmetic series: n/2 * (first term + last term), where n is the number of terms. Plugging in the values, we get 20/2 * (3 + 193) = 10 * 196 = 1960. Therefore, the sum of the series is 1960.
An arithmetic series is a fairly similar to an arithmetic sequence except for the fact that in a series you are adding the numbers in between, not putting commas. Example: Sequence 1,3,5,7,.........n Series 1+3+5+7+..........+n Hope this helped(:
The given series is known as a sequence of perfect cubes with alternating signs. The rule for this series is that each term is the cube of the next integer in the sequence, alternating between positive and negative. In this case, the series starts with -1, then continues with 2^3 = 8, -3^3 = -27, 4^3 = 64, and -5^3 = -125.
For any number n, the numbers fit the formula (n-1)3+(n-1)2+(n-1). Therefore, the seventh number in the series is equal to 63+62+6 = 216 + 36 + 6 = 258.
1875, 9375, 46875The pattern is:3 x 5(n-1)Where n denotes the nth number in the series.
30. The series can be generated by: t(n) = (-n3 + 3n2 - 56n + 510)/6 for n = 1, 2, 3, ...