When the odd number term of a sequence is 0, the even number term is infinitely close to 0. Is it infinitely close to 0? {1 + (- 1) nth power / 2 * 1 / N}, divergence? Convergence
Of course
{(1/n)·[(1+(-1)^n]/2}
That is, 0,1 / 2,0,1 / 4,..., of course, it is convergent and easy to verify that it satisfies the definition
RELATED INFORMATIONS
- 1. The first n terms of sequence {an} and the n-th power of Sn = a (a is a nonzero constant) are known Then [an] is the sequence of equal ratio; it is the sequence of equal difference; it may be equal difference or equal ratio,
- 2. Comparison: SN = 1 / 2 + 2 / 4 + 3 / 8 + 4 / 16 +The nth power of N / 2 (n is a positive integer) is clearly related to the size of 2 I don't understand every step, especially subtraction
- 3. Compare Sn = 12 + 24 + 38 + 416 + +N2n (n is any natural number) and the size of 2______ .
- 4. Comparison: SN = 1 / 2 + 2 / 4 + 3 / 8 + 4 / 16 +The power of N / 2 (n is a positive integer) and the size of 2
- 5. Let n be a positive integer and n ^ 2 + 1085 be a positive integer power of 3
- 6. The nth power of 2.8 * the nth power of 16 = the 22nd power of 2 to find the value of positive integer n
- 7. Given that n is a positive integer and the nth power of (the 2nd power of x) is 9, find the 2nth power of the 2nd power of (- 1 / 3 times the 3nth power of x) - 3 (- the 2nd power of x)
- 8. The number of digits of the nth power of a positive integer The number of digits of positive integer power is regular. When the number of digits of positive integer a is 0,1,5,6, the number of digits of a ^ n is still 0,1,5,6; when the number of occurrences of a is 4,9, the number of digits of power will appear repeatedly every time the exponent of a increases by 2; when the number of digits of a is 2,3,7,8, the number of digits of power will appear repeatedly every time the exponent increases by 4 If the single digit of a ^ k is a, then the last digit of a ^ 4m + k is also a (k is a positive integer, M is a non negative integer) Example 1 find the number of 2003 ^ 2005 2003^2005=2003^4×501+1. Because the single digit of 2003 ^ 1 is the same as that of 3 ^ 1, the single digit of 2003 ^ 2005 is 3 Example 2 find the integer x satisfying the equation x ^ 5 = 656356768 ∵ 10 ^ 5 is 6 digits. 100 ^ 5 = 10 ^ 10 is 11 digits Because x ^ 5 is 9 digits. So 10
- 9. How many positive integers is it to find the 12 power of n = 2 multiplied by the 8 power of 5?
- 10. Fill in the blanks with the code: the coefficients of each item in the expansion of the nth power of (a + b) are very regular. These coefficients form the famous Yang Hui triangle The following program gives the calculation method of the nth coefficient of the m-th layer, try to perfect it (m, n are calculated from 0) \x05public static int f(int m,int n) \x05{ \x05\x05if(m==0) return 1; \x05\x05if(n==0 || n==m) return 1; \x05\x05return __________________________ ;
- 11. What is the sum of all terms of n-1 power of finite sequence 1,1 + 2,1 + 2 + 4.1 + 2 + 4 +... + 2?
- 12. Finding the sum of the first n terms of the sequence {(n + 1) / 2}
- 13. Find the square of sequence 1, negative 2, negative 4 N of (negative 1) minus the square of 1 power n What's the difference Find the square of sequence 1, negative 2, negative 4 N of (negative 1) minus the square of 1 power n And. Urgent
- 14. How to make the sequence positive and negative Alternating like this? (expressed with the power of - 1) The answer is n (n-1) / 2 of (- 1), n = 0,1,2 Why?
- 15. Negative power column 1 4 3 1 / 5 1 / 36(
- 16. Given the square of X - the square of y = 12, x + y = 6, find the value of Y power of X - x power of Y
- 17. It is known that: the x power of 12 = 3, the Y power of 12 = 2, find 1-2y / 1-x + y of 8 It is known that: the x power of 12 = 3, the Y power of 12 = 2, find 1-2y / 1-x + y of 8
- 18. The 1000 power of 3
- 19. If positive numbers a and B satisfy 2A + 3B = 6, then the minimum value of 2 / A + 3 / B is
- 20. Given 2A + 3B = 6, find the minimum value of a ^ 2 + B ^ 2? Help to use the basic inequality