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Help me evaluate the infinite series of

\[\displaystyle 1 + \sqrt {2 + \sqrt[3] {3+ \sqrt[4] {4 + \sqrt[5] {5 +......}}}}\].

Note by Bryan Lee Shi Yang
2 years, 8 months ago

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Thanks for all your suggestions. I will post it into a question.

Bryan Lee Shi Yang - 2 years, 8 months ago

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Then does your number have a symbol, I mean Greek symbol or anything?

Bryan Lee Shi Yang - 2 years, 8 months ago

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1
2
3
4
5
6
>>>import math
>>>ans = 0.0
>>>for x in xrange(10, 0, -1):
>>>    ans = math.pow((x + ans), 1.0/x)
>>>print("%.50f" % (ans))
2.91163921624582400227154721505939960479736328125000

Using \(12\) or above to starts iteration gives equal amount precision, about \(50\) digits.

Damiann Mangan - 2 years, 8 months ago

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Same method, but did it manually.

Archit Boobna - 2 years, 8 months ago

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Awesome question!

John Young Staff - 2 years, 8 months ago

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please solve and post a nice solution to this

Ritam Baidya - 2 years, 8 months ago

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The only thing I know is that it is less than 3, because of an identity by Ramanujan.Since it is above 2.9, we can find a good approximation.I dont think a closed form exists.

Bogdan Simeonov - 2 years, 8 months ago

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Ramanujan maybe?

Krishna Ar - 2 years, 8 months ago

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I think that the answer is about 2.911

Archit Boobna - 2 years, 8 months ago

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And also give full step solutions (as much as possible) , bcs. I haven't figured out yet!

Bryan Lee Shi Yang - 2 years, 8 months ago

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Wait....you changed the problem -_- I was working on the previous one :V

Krishna Sharma - 2 years, 8 months ago

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Haha XD

Bryan Lee Shi Yang - 2 years, 8 months ago

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I don't have any solution but out of curiosity, just to find out, I approximated it and got 2.911.

I made an equation ((((x-1)^2-2)^3-3)^4-4)^4........=x which gives 2.911

Archit Boobna - 2 years, 8 months ago

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Please enter the answer correct to 3 decimal places.

Bryan Lee Shi Yang - 2 years, 8 months ago

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