1 5 3 2 2 8 8 . . .
The n th term of the above nested radical is given by 1 ∑ n x x . Determine if the nested radical converges or diverges and than, if it converges, estimate its value k .
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@Cristiano Sansó , i dont understand the very first line. Can u elaborate how the nested radical can be written like that?
each term of the radical multiplies all the other terms
imagine seeing it like
1 5 3 2 . . . =
= 2 1 1 2 2 5 2 3 3 2 . . .
After slowing down the exponential growth, you assumed it will be less than 8. Where is the proof for that?
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this is the reason I wrote estimate and not approximate... anyway when you see a growth for n
2 ; 2 ; 2 ; 3 ; 5 ; . . .
since an exponential growth can't decrease, the next number must be at least 7, the next one at least 10, the next one at least 14, 20, 28 and so on
so we expect something much bigger than
2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 1 0 ; 1 4 ; 2 0 ; 2 8 ; . . .
so considering that I slowed the exponential growth down to
2 ; 2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 1 0 ; 1 4 ; 2 0 ; 2 7 ; 3 8 ; 5 6 ; . . .
it's so unlikely that k will exceed that value.
If you compare
2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 1 0 ; 1 4 ; 2 0 ; 2 8 ; . . .
2 ; 2 ; 2 ; 2 ; 3 ; 5 ; 7 ; 1 0 ; 1 4 ; 2 0 ; . . .
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first of all, let's determine if it converges. We can see, the nested radical can be written as an infinite product.
n = 1 ∏ ∞ 2 n x = 1 ∑ n x x
An infinite product
n = 1 ∏ ∞ a n
converges if
n → ∞ lim a n = 1 (at least for nested radicals).
In this case, we need
n → ∞ lim 2 n x = 1 ∑ n x x = 1
in that sum, the highest value is for x = n and all the other terms are negligible, so,
2 n x = 1 ∑ n x x ∼ 2 n n n
for n → ∞
This is the way I solved this limit
n → ∞ lim ( n n ) 2 n 1 = n → ∞ lim n 2 n n = n → ∞ lim e 2 n n ln ( n ) = e 0 = 1
So, now we know the product converges, let's see how to estimate the value.
let's consider the first few terms of the nested radical
1 ; 5 ; 3 2 ; 2 8 8 ; 3 4 1 3 ; 5 0 0 6 9
so, the value of the radical (k)
k > 4 5 ∗ 8 3 2 ∗ 1 6 2 8 8 ∗ 3 2 3 4 1 3 ∗ 6 4 5 0 0 6 9
we can compare each of these to, for example, n 2 so,
4 5 > 2
8 3 2 > 2
1 6 2 8 8 > 2
3 2 3 4 1 3 > 3 2
6 4 5 0 0 6 9 > 5 2
and than
4 < 2 ∗ 2 ∗ 2 ∗ 3 2 ∗ 5 2 < k
from the inequalities above, we can expect an exponential growth of n in n 2 , like
2 ; 2 ; 2 ; 3 2 ; 5 2 ; 7 2 ; 1 1 2 ; 1 8 2 ; 3 1 2 ; 5 4 2 ; 9 6 2 ; . . .
To estimate a maximum value we can try to slow down the exponential growth, like
2 ; 2 ; 2 ; 2 ; 3 2 ; 5 2 ; 7 2 ; 1 0 2 ; 1 4 2 ; 2 0 2 ; 2 7 2 ; 3 8 2 ; 5 6 2 ; . . .
and now we can say
4 < k < 2 ∗ 2 ∗ 2 ∗ 2 ∗ 3 2 ∗ 5 2 ∗ 7 2 ∗ 1 0 2 ∗ 1 4 2 ∗ 2 0 2 ∗ 2 7 2 ∗ 3 8 2 ∗ 5 6 2 < 8
and finally
4 < k < 8
We can check with a calculator, to find out that the value of the nested radical is 6 . 5 2 3 0 6 1 7 6 . . .