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### Tricks to Sum an Infinite Series

```
Date: 01/22/99 at 22:38:20
From: Steve Heggie
Subject: Convergence of a strange sum

infinity
SUM    [((-1)^n)*|sin n|]/n
n = 1

Do you have any ideas on how to find the convergence of this series?
I'm pretty sure it converges on zero, but I cannot begin to prove it
(aside from just adding up the partial sums).

Thanks,
Steve
```

```
Date: 01/23/99 at 08:19:13
From: Doctor Anthony
Subject: Re: Convergence of a strange sum

The series is  -sin(1) + sin(2)/2 - sin(3)/3 + .......

= -IM[e^(i) - e^(2i)/2 + e^(3i)/3 - .....

where IM means 'imaginary part of'. Let e^i = x and the series can be
written:

= -IM[x - x^2/2 + x^3/3 - x^4/4 + .......

Differentiating the series we get:

= -IM[1 - x + x^2 - x^3 + .......

= -IM[1/(1+x)]

To find the sum of the actual series we must integrate this, to get:

= -IM[ln(1+x)]

= -IM[ln(1+e^i)]

Suppose ln(1+e^i) = p+iq and we require the value of -q. Then:

1 + e^i = e^(p+iq)

1 + e^i = e^p e^(iq)

1 + cos(1) + isin(1) = e^p[cos(q) + isin(q)]

1 + cos(1) + isin(1) = e^p cos(q) + i e^p sin(q)

Therefore:

1 + cos(1) = e^p cos(q)
sin(1) = e^p sin(q)

Dividing these equations:

tan(q) = sin(1)/[1+cos(1)]

q = tan^(-1)[sin(1)/(1+cos(1))] = 0.5

We require -q as the sum of the given series, so the required sum is
-1/2.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/23/99 at 16:23:39
From: Steven Heggie
Subject: Re: Convergence of a strange sum

Dr. Anthony,

I've been working through the solution that you provided to my
infinite sum problem ((-1)^n * |sin (n)|) / n, and I have figured
everything out except for one part. It is the line that suggests that
i sin(1) = i e^p sin(q). I don't understand why. Also, can you explain
the line right after that?

Thank you,
Steve
```

```
Date: 01/23/99 at 17:49:55
From: Doctor Anthony
Subject: Re: Convergence of a strange sum

When you have a complex equation like:

a + ib = p + iq

then we can equate the real part on the left with the real part on the
right, that is, a = p, and equate the imaginary part on the left with
the imaginary part on the right, that is, b = q.

It is easy to prove this. Write the equation:

a-p = i(q-b)

Now square both sides:

(a-p)^2 = -(q-b)^2    since i^2 = -1

(a-p)^2 + (q-b)^2 = 0

This requires the sum of two positive terms to equal 0. Clearly this is
impossible and in fact each bracket must separately equal 0. That is,
a = p and b = q.

Applying this idea to the equation:

1 + cos(1)+i sin(1) = e^p cos(q) + i e^p sin(q)

We must have:

1 + cos(1) = e^p cos(q)    equating real part on each side
sin(1) = e^p sin(q)        equating imaginary part on each side

I hope that has cleared up the difficulty.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/23/99 at 18:02:35
From: Steven Heggie
Subject: Re: Convergence of a strange sum

Dr. Anthony,

Thanks for the help. I NEVER would have figured this problem out on my
own. Thank goodness for Euler's Formula!

- Steve Heggie
```
Associated Topics:
High School Sequences, Series

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