You need to find a balance between the two.

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, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

 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.

- 3 years, 5 months ago

Same method, but did it manually.

- 3 years, 5 months ago

Awesome question!

Staff - 3 years, 5 months ago

please solve and post a nice solution to this

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

Ramanujan maybe?

- 3 years, 5 months ago

I think that the answer is about 2.911

- 3 years, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Haha XD

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Please enter the answer correct to 3 decimal places.

- 3 years, 5 months ago