Proof of alternating series test, Mathematics

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Proof of Alternating Series Test

With no loss of generality we can assume that the series begins at n =1. If not we could change the proof below to meet the new starting place or we could perform an index shift to obtain the series to begin at n =1 .

First, notice that because the terms of the sequence are decreasing for any two successive terms we can say,

bn - bn+1 ≥ 0

Here now, let us take a look at the even partial sums.

s2 = b1 - b2 ≥ 0

s4 = b1 - b2 + b3 - b4 = s2 + b3 - b4 ≥ s2                                              because b3 - b4 > 0

S6 = s4 + b5 - b6  ≥ s4                                                            because b5 - b6 > 0     

S2n = S2n -2 + b2n -1 - b2n  ≥ S2n -2                                                           because b2n-1 - b2n > 0

Thus, {S2n}is an increasing sequence.

 Next, we can as well write the general term as,

S2n = b1-b2 + b3 - b4 + b5 + .... - b2n-2 + b2n-1 - b2n

= b1 - (b2-b3) - (b4 - b5) + ..... - (b2n-2 - b2n-1) - b2n

Every quantity in parenthesis is positive and by assumption we be familiar with that b2n is as well positive.  Thus, this tells us that S2n< b1 for all n.

We now be familiar with that {S2n}is an increasing sequence that is bounded above and thus we know that it must as well converge.  Thus, let's assume that its limit is s or,

1578_Proof of Alternating Series Test 1.png

Subsequently, we can quickly find out the limit of the sequence of odd partial sums, {S2n+1} as follows,

1043_Proof of Alternating Series Test 2.png

Thus, we now know that both of the {S2n} and {S2n+1} are convergent sequences and they both have similar limit and so we as well know that {Sn} is a convergent sequence along with a limit of s.  This in turn tells us that ∑an is convergent.


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