Alternating Series

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An alternating series is a series whose terms are alternately positive and negative. Namely, it has the form

sum((-1)^n*b[n],n = 1 .. infinity) or sum((-1)^(n+1)*b[n],n = 1 .. infinity) where 0 <= b[n] .

Here are two examples :

1. sum((-1)^n*n/(n+1),n = 1 .. infinity) = -1/2+2/3-3/4+4/5-5/6 + ... .

2. sum(cos((n+1)*Pi),n = 1 .. infinity) = 1-1+1-1+1 + ... .

3. sum((-1)^(n+1)/n,n = 1 .. infinity) = 1-1/2+1/3-1/4+1/5-1/6 + ... .

For the first and second examples, since limit(cos((n+1)*Pi),n = infinity) <> 0 and limit((-1)^n*n/(n+1),n = infinity) <> 0 , by The Test for Divergence , both series diverge.

Let's take a look at the third example sum((-1)^(n+1)/n,n = 1 .. infinity) = 1-1/2+1/3-1/4+1/5-1/6 + ... .

Here we compute the nth partial sum S[n] for n = 1 , 2, ... , 10 .

s := proc (n) options operator, arrow; sum((-1)^(k+...

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S[1] := 1.

S[2] := .5000000000

S[3] := .8333333333

S[4] := .5833333333

S[5] := .7833333333

S[6] := .6166666667

S[7] := .7595238095

S[8] := .6345238095

S[9] := .7456349206

S[10] := .6456349206

S[11] := .7365440115

S[12] := .6532106782

S[13] := .7301337551

S[14] := .6587051837

S[15] := .7253718504

S[16] := .6628718504

S[17] := .7216953798

S[18] := .6661398242

S[19] := .7187714032

S[20] := .6687714032

We have better picture of the series from the plot below.


[Maple Plot]

It seems that { S[2*n-1] } is decreasing and { S[2*n] } is increasing. It looks more clearly in the following plot, where { S[2*n-1] } is plotted with red circles and { S[2*n] } is plotted with blue circles.


[Maple Plot]

We can see from the graph above that { S[2*n] } is increasing and bounded above by a[1] , while { S[2*n-1] } is decreasing and bounded below by a[2] ; by the monotonic Sequence Test, both { S[2*n] } and { S[2*n-1] } converge.

Moreover,

S[2*n] = S[2*n-1]-1/(2*n)

it follows that limit(S[2*n],n = infinity) = limit(S[2*n-1],n = inf... . Therefore, limit(S[n],n = infinity) converges.

In general, we have the following theorem :

The Alternating Series Test

If 0 <= b[n] and { b[n] } is decreasing with limit(b[n],n = infinity) = 0 , then the alternating series

sum((-1)^(n+1)*b[n],n = 1 .. infinity) ( or sum((-1)^n*b[n],n = 1 .. infinity) )

converges.

In fact,

S[2*n] = sum((-1)^(k+1)*b[k],k = 1 .. 2*n) = ( b[1]-b[2] ) + ( b[3]-b[4] ) + ... + ( b[2*n-1]-b[2*n] )

since { b[n] } is decreasing, we have 0 <= S[2*n] and { S[2*n] } is decreasing.

But we can also write

S[2*n] = b[1]-(b[2]-b[3]) - ... - ( b[2*n-2]-b[2*n-1] ) - b[2*n]

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so S[2*n] <= b[1] for all n .

Therefore, by the Monotonic Sequence Theorem, { S[2*n] } converges, say limit(S[2*n],n = infinity) = S .

Moreover, S[2*n+1] = S[2*n]+b[2*n+1] and limit(a[n],n = infinity) = 0 , it follows that { S[2*n+1] } converges and

limit(S[2*n+1],n = infinity) = limit(S[2*n],n = inf... = S

Therefore, limit(S[n],n = infinity) = S .

Note that S <= b[1] (Why ?).

In this case,

abs(S-S[n]) = abs(sum((-1)^(k+1)*b[k],k = n+1 .. in... = b[n+1]-b[n+2]+b[n+3] + ... + ....

For each 1 <= n , sum((-1)^(n+1)*b[n+k],k = 1 .. infinity) = b[n+1]-b... + ... + .... is also an alternating series which satisfies the assumption of the the Alternating Series Test. The following plot illustrates the case for n = 40 ( the blue circle represents the point ( 41, b[41] ) ).

[Maple Plot]

It is therefore convergent, let's call the limit R[n] , then we have

abs(S-S[n]) = R[n] <= b[n+1] .

We say the series sum(a[n],n = 1 .. infinity) is absolutely convergent if the series of absolute values sum(abs(a[n]),n = 1 .. infinity) converges.

For example, sum((-1)^(n+1)/(n^2),n = 1 .. infinity) .

By the Integral Test, sum(abs((-1)^(n+1)/(n^2)),n = 1 .. infinity) = sum(... converges, so sum((-1)^(n+1)/(n^2),n = 1 .. infinity) is absolutely convergent; and by the Alternating Series Test, we get that sum((-1)^(n+1)/(n^2),n = 1 .. infinity) is convergent also.

Note that the series sum((-1)^(n+1)/n,n = 1 .. infinity) converges, but sum(abs((-1)^(n+1)/n),n = 1 .. infinity) = sum(1/n,... diverges. From this, we see that a convergent series may not be absolutely convergent.

Question :

Is it true that if sum(a[n],n = 1 .. infinity) is absolutely convergent then sum(a[n],n = 1 .. infinity) is convergent ?

Let's examine the difference between the convergence of the series sum((-1)^(n+1)/n,n = 1 .. infinity) and that of the series sum((-1)^(n+1)/(n^2),n = 1 .. infinity) .

S1 := proc (n) options operator, arrow; sum((-1)^(k...

S2 := proc (n) options operator, arrow; sum((-1)^(k...

[Maple Plot]

It seems sum((-1)^(n+1)/(n^2),n = 1 .. infinity) converges much faster than sum((-1)^(n+1)/n,n = 1 .. infinity) .

Now let's compare the convergence of the series sum(1/(n^2),n = 1 .. infinity) and sum((-1)^(n+1)/(n^2),n = 1 .. infinity) .

S1 := proc (n) options operator, arrow; sum((-1)^(k...

S3 := proc (n) options operator, arrow; sum(1/(k^2)...

[Maple Plot]

The rate of convergence of these two series are about the same.

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[Maple Plot]

From the graphs above, we see that the absolute value of the n th term of the series sum((-1)^(n+1)/(n^2),n = 1 .. infinity) goes to zero more rapidly than that of sum((-1)^(n+1)/n,n = 1 .. infinity) .

It seems that we have a good reason to conjecture that if sum(a[n],n = 1 .. infinity) is absolutely convergent then sum(a[n],n = 1 .. infinity) is convergent.

Theorem If sum(a[n],n = 1 .. infinity) is absolutely convergent then sum(a[n],n = 1 .. infinity) is convergent.

Proof :

Note that 0 <= a[n] + abs(a[n]) <= 2*abs(a[n]) .

If sum(a[n],n = 1 .. infinity) is absolutely convergent then sum(abs(a[n]),n = 1 .. infinity) is convergent. So sum(2*abs(a[n]),n = 1 .. infinity) is convergent. Therefore, by Comparison Test,

sum(a[n]+abs(a[n]),n = 1 .. infinity) is convergent.

Since a[n] = [a[n]+abs(a[n])]-abs(a[n]) , so sum(a[n],n = 1 .. infinity) is the difference of two convergent series and is therefore convergent.

We say a series sum(a[n],n = 1 .. infinity) converges conditionally iff sum(a[n],n = 1 .. infinity) converges but sum(a[n],n = 1 .. infinity) does not converge absolutely.

Now we have tests for positive terms and for alternating series. But what if the sign of the terms switches back and forth irregularly ? For example, sum(cos(n)/n,n = 1 .. infinity) ?

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S[1] := .5403023059

S[2] := -.2080734182

S[3] := -.3299974988

S[4] := -.1634109052

S[5] := .5673243710e-1

S[6] := .1600283811

S[7] := .1077003221

S[8] := -.1818750422e-1

S[9] := -.1012366958

S[10] := -.8390715291e-1

S[11] := .4023361807e-3

S[12] := .7032116322e-1

S[13] := .6980359857e-1

S[14] := .9766944157e-2

S[15] := -.5064586086e-1

S[16] := -.5985371752e-1

S[17] := -.1618607871e-1

S[18] := .3668426157e-1

S[19] := .5203708517e-1

S[20] := .2040410309e-1

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[Maple Plot]

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[Maple Plot]

We have no clue !!!

sum(cos(n)/n,n = 1 .. infinity)

sum(cos(n)/n,n = 1 .. infinity)

But it is rather different for the case if we consider the series sum(cos(n)/(n^2),n = 1 .. infinity) .

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[Maple Plot]


[Maple Plot]

The series seems to converge !! We notice that the terms of the series sum(cos(n)/(n^2),n = 1 .. infinity) converges to 0 much faster than that of the series sum(cos(n)/n,n = 1 .. infinity) and also sum(cos(n)/(n^2),n = 1 .. infinity) converges absolutely (why ??).

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[Maple Plot]

Something funny about the plot for n <= 10 --- It does not seem to be increasing ! Let's take a closer look.


[Maple Plot]

It's O.K. It is increasing. But can we be sure that sum(cos(n)/(n^2),n = 1 .. infinity) converge ?

Yes, since sum(cos(n)/(n^2),n = 1 .. infinity) converges absolutely, by the theorem, sum(cos(n)/(n^2),n = 1 .. infinity) converge.

Remark : If sum(abs(a[n]),n = 1 .. infinity) converges, by the theorem, sum(a[n],n = 1 .. infinity) converges. However, in general, sum(abs(a[n]),n = 1 .. infinity) <> sum(a[n],n = 1 ... . Examples ???

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Download Maple Worksheet

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