EXPLAIN PLEASE BECAUSE I NEED TO KNOW HOW TO DO THIS Let Sk

EXPLAIN PLEASE BECAUSE I NEED TO KNOW HOW TO DO THIS.

Let Sk be the sum of the first k terms of the alternating series The series has positive decreasing terms an whose limit is 0, so you know it converges. Suppose you have two partial sums Sj and Sm, where j

Solution

One way to approach this problem is to examine a real series where a_n = (1/2)ⁿ.

The quantity |S_j - S_m| is the tail obtained when subtracting one partial sum from another.

If j = 3 and m = 6, the tail is | - (1/16) + (1/32) - (1/64) | = 3/64

The first term of the tail, a_j+1, is greater than the tail, if it\'s even. Because the series alternates, if j is even, a_j+1 will be negative and less than the tail. Choice A is eliminated because it is only greater than the tail under certain circumstances.

Keeping in mind that the tail is a sequence of consecutive terms from S, a_m+1 is the next term. When a_m+1 is negative, it is necessarily less than the tail, which is positive due to the absolute value of the difference between S_j and S_m. For (1/2)ⁿ, a_m+1 is also less than the difference between consecutive pairs of terms in the tail, it is necessarily smaller than the sum of those differences, which is the tail. This eliminates choice B.

That leaves the two choices that are series.

Let h be the difference between m and j. The integer h corresponds to the number of elements in the tail. S_m-j+1 is the partial sum comprised of the first h + 1 elements. Because the elements decrease, the partial sum of the first h elements is generally going to be larger than the tail. Consider very short sequences: m = 2 and j = 1. For (1/2)ⁿ, S_j = 1/2 and S_m = 1/2 - 1/4 = 1/4. The tail, |1/4 - 1/2| = 1/4. S_m-j+1 = S₂ = 1/2 - 1/4 = 1/4. For an alternating harmonic series, which converges more slowly than (1/2)ⁿ, the partial sum of the first h terms will exceed the tail. If the series converges more precipitately than (1/2)ⁿ the sum of the first h terms should be less than the tail. < I\'m asserting this without proof.> This leaves out choice C.

For the last case, consider the following mappings in the partial series S_j. Where a_j is positive, it is mapped onto a₁, a_j+1 onto a₂, until all elements of the tail are mapped onto the first m - j elements of S_j. Because the series decreases, the initial terms of S_j exceed the terms of the tail they\'re mapped onto. The unmapped terms upto a_j+1 gently increase the difference so that S_j+1 is always greater than the tail. For a_j negative, it is mapped onto a₂, et cetera, with a₁ and the other unmapped terms increasing the difference between the tail and S_j+1.