**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,

b_{n} - b_{n+1} ≥ 0

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

s_{2} = b_{1} - b_{2} ≥ 0

s_{4 }= b_{1} - b_{2} + b_{3} - b_{4} = s_{2} + b_{3} - b_{4} ≥ s_{2 }because b_{3} - b_{4} > 0

S_{6} = s4 + b5 - b6 ≥ s4 because b_{5} - b_{6} > 0

S_{2n} = S_{2n -2} + b_{2n -1} - b_{2n} ≥ S_{2n -2 }because b_{2n-1} - b_{2n} > 0

Thus, {S_{2n}}is an increasing sequence.

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

S_{2n} = b_{1}-b_{2} + b_{3} - b_{4} + b_{5} + .... - b_{2n-2} + b_{2n-1} - b_{2n}

= b_{1} - (b_{2}-b_{3}) - (b_{4} - b_{5}) + ..... - (b_{2n-2 }- b_{2n-1}) - b_{2n}

Every quantity in parenthesis is positive and by assumption we be familiar with that b_{2n }is as well positive. Thus, this tells us that S_{2n}< b_{1} for all n.

We now be familiar with that {S_{2n}}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,

Subsequently, we can quickly find out the limit of the sequence of odd partial sums, {S_{2n+1}} as follows,

Thus, we now know that both of the {S_{2n}} and {S_{2n+1}} are convergent sequences and they both have similar limit and so we as well know that {S_{n}} is a convergent sequence along with a limit of s. This in turn tells us that ∑a_{n }is convergent.