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summation of infinite series

how to solve this? it is question 8
(edited 7 years ago)
20160731_122857[1].jpg240.9KB
Original post by Akhito Kanbara
how to solve this? it is question 8


I have an idea, got the answer as 1. The trick is to notice the pattern and use some substitutions. Before I post my method, what have you tried though?
(edited 7 years ago)
Original post by Akhito Kanbara
how to solve this? it is question 8


The trick to these sorts of questions is to realise its far to complicated to actually evaluate them.You need to use the fact that these expressions are truly infinite and thus you can rewrite them(effectively the same as applying certain functions to them) without changing them and thus use this to write an algebraic equation for those expressions for the top and bottom and divide the answers you find.
(edited 7 years ago)
Original post by Akhito Kanbara
how to solve this? it is question 8


What happens if you cube the expression?
I tried them to see the pattern for geometric progression i dun think it has

Posted from TSR Mobile
This wat i reach use dalik way

Posted from TSR Mobile
1469959214426.jpg28.2KB
Original post by Akhito Kanbara
This wat i reach use dalik way

Posted from TSR Mobile


That doesn't lead you anywhere, or perhaps I don't follow.

Try this:
Let x=444...333x=\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}
Then by definition x=4x3x=\sqrt[3]{4x}

Solve for xx and disregard any solutions which do not fit this (such as 0).

Then let y=222...y=\sqrt{2\sqrt{2\sqrt{2...}}} and apply the same trick and solve for yy while disregarding any solutions which don't fit in.

If you observe, you final answer should be xy\frac{x}{y}
Reply 7
Avatar for Zacken
Zacken
22
Bah, I don't agree with questions like these, you need to prove that the sequence xn+1=2xnx_{n+1} = \sqrt{2 x_{n}} converges for you to make any reasonable use of the techniques above. Otherwise you can come up with some truly nonsensical results.
Reply 8
Avatar for B_9710
B_9710
18
You could define a sequence say xn+1=4xn3,x1=1 x_{n+1} = \sqrt[3]{4x_n}, x_1 =1 . You should know how to find the limit of the sequence (given that it converges - no proof needed) as its basic C2.
You can do a similar thing for the denominator. And the answer you need will be lim⁡n→∞xnyn \displaystyle \lim_{n\to\infty} \frac{x_n}{y_n} .
(edited 7 years ago)
Reply 9
Avatar for B_9710
B_9710
18
Original post by Zacken
Bah, I don't agree with questions like these, you need to prove that the sequence xn+1=2xnx_{n+1} = \sqrt{2 x_{n}} converges for you to make any reasonable use of the techniques above. Otherwise you can come up with some truly nonsensical results.


Yes, it will have people believe that un+1=2un+1 \text{u}_{n+1}=2\text{u}_n+1 is 1.
Reply 10
Avatar for Zacken
Zacken
22
Original post by B_9710
Yes, it will have people believe that un+1=2un+1 \text{u}_{n+1}=2\text{u}_n+1 is 1.


Or that
Unparseable latex formula:

\displaystyle x^{x^{x}^{\cdots}} = 2

and
Unparseable latex formula:

\displaystyle x^{x^{x}^{\cdots}} = 4

converges to the same xx. Hitch is that only the first one is right since it only converges in [e−1,e][e^{-1}, e] or something similar.

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