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### Alternating Harmonic Series

```
Date: 11/18/97 at 14:18:46
From: Phil Campbell
Subject: Alternating harmonic series

I am desperately trying to find the proof for the sum of the
alternating harmonic series. I did find out that it is ln(2), but
please tell me why. We discussed it in calculus for two days after I
brought it up, but couldn't even figure out what it converges to.
I looked it up and found out, but I need the proof so I can show it to
other curious parties, and so I can once again sleep at night.

Thanks.
```

```
Date: 11/18/97 at 15:49:01
From: Doctor Pete
Subject: Re: Alternating harmonic series

Hi,

Here is the proof. It requires an understanding of Taylor series.

Theorem:  Sum[(-1)^(n-1)/1, {n,1,Infinity}] = 1 - 1/2 + 1/3 - ... =
Ln[2].

Proof:  Recall that the Taylor series expansion of a function F[x]
about a point x = a is

F[x] = Sum[F^(n)[a]*(x-a)^n/n!,{n,0,Infinity}],

where F^(n)[a] is the n(th) derivative of F[x] with respect to x,
evaluated at x = a.  When a = 0, we have the Maclaurin expansion

F[x] = Sum[F^(n)[0]*x^n/n!,{n,0,Infinity}].

Now, let F[x] = Ln[1+x].  Then the successive derivatives with respect
to x are

F'[x] = (1+x)^(-1)
F''[x] = -(1+x)^(-2)
F'''[x] = 2(1+x)^(-3)
F^(4)[x] = -6(1+x)^(-4)
...
F^(n)[x] = (-1)^(n-1)(n-1)!(1+x)^(-n), n > 0.

since we see the sign alternating (positive when we take an odd
derivative, negative when we take the even derivative).  At x = a = 0,
we find

F[0] = Ln[1+0] = 0,
F^(n)[0] = (-1)^(n-1)(n-1)!, n > 0.

Therefore, the n(th) term in the Maclaurin series expansion of Ln[1+x]
is

F[0]*x^0/0! = 0*1 = 0,
F^(n)[0]*x^n/n! = (-1)^(n-1)*x^n/n, n > 0.

since (n-1)!/n! = 1/n.  So it follows that

Ln[1+x] = Sum[(-1)^(n-1)*x^n/n,{n,1,Infinity}]

(note the index begins on 1, not 0, because the term where n = 0 is
0).  But if we let x = 1 in the above, we obtain

Ln[2] = Sum[(-1)^(n-1)/n,{n,1,Infinity}],

which is what was to be proved.
________

Note that the interval of convergence of the Maclaurin series for
Ln[1+x] is -1 < x <= 1; that is, any value outisde of this range
will not be a valid convergent series. To see why, we apply the
ratio test and obtain |x| < 1; but as you have proved, the alternating
harmonic series, where x = 1, does converge. Therefore the interval
includes this endpoint. Obviously, when x = -1, Ln[1+x] = Ln[0]
and is not defined, but observe that the term in the series is then
(-1)^(n-1)*(-1)^n/n = -1/n; hence Sum[1/n,{n,1,Infinity}] = -Ln[0],
which does not converge! (Of course, you can't really say this - it
is because the harmonic series fails to converge that the Maclaurin
series fails to be valid at that point, not the other way around.)
Finally, we can prove lots of other interesting sums with this series,
for instance

1/2 - 1/8 + 1/24 - 1/64 + ... = Ln[1+1/2] = Ln[3/2] = Ln[3] - Ln[2].

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 11/18/97 at 15:52:24
From: Doctor Wilkinson
Subject: Re: Alternating harmonic series

ln(2) is the integral from 1 to 2 of dx/x.  Make the substitition
x = 1 + t to transform this to the integral from 0 to 1 of dt/(1+t).
Expand 1/(1+t) into the geometric series 1 - t + t^2 - t^3 +-... and
integrate term-by-term.  (Of course you have to justify the term-by-
term integration.)

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis
High School Calculus
High School Sequences, Series

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