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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 3 years, 2 months ago

Easy Math Editor

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**bold**

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[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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

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

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.

Same method, but did it manually.

Awesome question!

please solve and post a nice solution to this

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.

Ramanujan maybe?

I think that the answer is about 2.911

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

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

Haha XD

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

Please enter the answer correct to 3 decimal places.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

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

Log in to reply

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

Log in to reply

Same method, but did it manually.

Log in to reply

Awesome question!

Log in to reply

please solve and post a nice solution to this

Log in to reply

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.

Log in to reply

Ramanujan maybe?

Log in to reply

I think that the answer is about 2.911

Log in to reply

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

Log in to reply

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

Log in to reply

Haha XD

Log in to reply

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

Log in to reply

Please enter the answer correct to 3 decimal places.

Log in to reply