# Does anyone know an infinite series summation that is equal to i?

The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?

Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i

Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i
If that works then it is clever and unexpected (a wonderfully typical trait of PF:) although it will work for anything in place of 'i' correct?
Being recursive is fine but I am looking for a 'unique' solution.

I have one and have a hard time believing I am first to do so, so what else is out there?

Yes, you may replace "i" with anything.

Yes, you may replace "i" with anything.
Borek posted on a couple of my previous threads 'make an infinite series equal to 1 and multiply (insert the value I was attempting to get) by it.' At least this is a cleverly disguised version of the same thing, well done!

Anyway, have you ever seen a unique infinite series equal to 'i'? Google searches and the such are turning up nothing. How cool would it be if this is a first?! Stephen Tashi
have you ever seen a unique infinite series equal to 'i'?
What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?

What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?
As in one where we don't use an infinite series equal to '1' and then multiply by 'i'. Apparently a fairly common suggestion here on PF :tongue2:

I do...but it's not an infinite sum...it's an infinite product.
Care to investigate this further? Or do you need a sum,
and can't see how a product could come in handy?

Because I have a product, ##log(I)## is a sum and is
easy to get. And powers of "I" are also easy to get
using the infinite product identity.

Would that be of any help in your project?

I do...but it's not an infinite sum...it's an infinite product.
Care to investigate this further? Or do you need a sum,
and can't see how a product could come in handy?

Because I have a product, ##log(I)## is a sum and is
easy to get. And powers of "I" are also easy to get
using the infinite product identity.

Would that be of any help in your project?
Honestly I have little experience with infinite products so far but I would love to see your work!

Honestly I have little experience with infinite products so far but I would love to see your work!

Given the information in the new thread "A Cool Identity" in General Math,
find an infinite number of infinite product identities for "I". Partial credit
will be given if you can only find a finite number of identities.
HINT: factor the factors into complex and real conjugates.

This is not a "snipe" hunt, we are dealing with the mathematical
certainty of prohibiting polynomial cross-products. Or stated
differently, an infinite number of equations in an infinite number
of unknowns, IS solvable.

Math is very fun.

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Stephen Tashi
As in one where we don't use an infinite series equal to '1' and then multiply by 'i'. Apparently a fairly common suggestion here on PF :tongue2:
Are you asking for a series that sums to i whose individual terms cannot each be divided by i ? If the individual terms can be divided by i, this gives an infinite series that sums to 1.

Honestly I have little experience with infinite products so far but I would love to see your work!
And I'll give you half credit if the only ones you find are an infinite number
of products equal to unity and trivially multiply by "I".

Math is very, very fun...or none of us would be here, n'est ce pas?

You don't have to be "super-brilliant", just interested in puzzles,
and then you become "super-brilliant" in some tiny, tiny niche of

EDIT: another HINT: be sure to remember that "I" need only be
multiplied by any one of the infinite number of factors.

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For those of you who do not like to cross threads,
here is the super-cool identity:

\begin{eqnarray}
\frac {1} {1-x} =
(1+x) \prod_{n=1} ^{\infty}
\left( \frac {1+x^{2^n}} {1-x^{2^n}} \right)^{2^{-n}} , {\;}
\text{for} {\;} 0\leq x<1
\nonumber
\end{eqnarray}

It's in LaTex, so feel free to move it around.

Just food for thought, which is better? Exposing "I", or hiding it?

Are you asking for a series that sums to i whose individual terms cannot each be divided by i ? If the individual terms can be divided by i, this gives an infinite series that sums to 1.
Sounds like we are on the same page, although you guys at PF are masters of semantics so I look forward to your response :)

ClamShell, I wish I could do as you are suggesting but infinite products are not yet part of my vocabulary.

Sorry to have imposed...I used to like sums because
their so easy to differentiate. Now I suspect that it
is just another fatal attraction. The answer is that
it is better to hide "I" than to expose it.
Disclaimer: But I do respect the opinion that "I"
is a pretty fun object. Adios.

Sorry to have imposed...I used to like sums because
their so easy to differentiate. Now I suspect that it
is just another fatal attraction. The answer is that
it is better to hide "I" than to expose it.
Disclaimer: But I do respect the opinion that "I"
is a pretty fun object. Adios.
Infinite products are the next step after infinite series, I can't wait to 'dig in' to these wonderful things. When I get up to speed I look forward to accepting your challenges!

Infinite products are the next step after infinite series, I can't wait to 'dig in' to these wonderful things. When I get up to speed I look forward to accepting your challenges!
Keep in touch.

The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?
Just can't keep my mouth shut...

What about 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno?)?

Each denominator is 2^n for n = 1 to infinity
or each denominator is double the previous denominator.

What happens if each denominator is (2i)^n ?

This simple sum may reveal what happens to
some class of other sums.

Or what happens when ##i = i/2 + i/4 + i/8 + ...## ?
and you need to find powers of ##i## by powering the
right hand side?

Just can't keep my mouth shut...

What about 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno?)?

Each denominator is 2^n for n = 1 to infinity
or each denominator is double the previous denominator.

What happens if each denominator is (2i)^n ?
Ah hah! With that we could do this,

$$i=.2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

I think that will work.

Very good, thank you ClamShell!
Anyone else have anything?

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Ah hah! With that we could do this,

$$i=.2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

I think that will work.

Very good, thank you ClamShell!
Anyone else have anything?

Hang on, I'm throwing it in wolfram...

$$i=2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

That's better, now it works So the sum is ##(2 - i)## ?

So the sum is ##(2 - i)## ?
Apparently so!

So what else can we do?