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Mentallic
Oct31-09, 05:31 AM
I'm curious to answer (or at least reasonably understand) what the answer to:

\sqrt{a}+\sqrt{a+\sqrt{a}}+\sqrt{a+\sqrt{a+\sqrt{a }}}+...

might be, where a>0.

It doesn't follow any ordinary pattern, such as an arithmetic or geometric progression. Also, if there is for any reason an easily derivable answer only for certain values of 'a', then that would also be interesting to hear.

edit: This problem diverges as hamster143 has noticed.
Instead, \sqrt{a}+\sqrt{a-\sqrt{a}}+\sqrt{a-\sqrt{a-\sqrt{a}}}+... where a>1

Hurkyl
Oct31-09, 05:35 AM
Have you experimented with it at all?

For example, looked at the first few terms for a chosen value of a?

Mentallic
Oct31-09, 05:47 AM
Yes I basically stared at the problem for a=2 and came up wth absolutely nothing. This is where I realized it's not a geometric sum.

The problem arose from when I was solving \sqrt{a+\sqrt{a+\sqrt{a}}}... and thought if I could expand it to this new problem. However, I'm out of luck.

hamster143
Oct31-09, 05:48 AM
Are you sure you got the expression right? As stated in your post, the sum diverges for all a except zero. Individual terms converge at (1+\sqrt{1+4a})/2.

Mentallic
Oct31-09, 06:00 AM
Oh it diverges, thanks for spotting that. A big wasted effort that was...

I think I'll change the question to:

\sqrt{a}+\sqrt{a-\sqrt{a}}+\sqrt{a-\sqrt{a-\sqrt{a}}}+..., a>1

hamster143
Oct31-09, 06:05 AM
That still diverges, terms converge at (-1+\sqrt{1+4a})/2.

Mentallic
Oct31-09, 06:09 AM
How frustrating. I can't even construct a problem that converges with the simple criteria that it be an infinite sum and the nth term having n nested surds among it.

I'll come back when I have a legitimate question.