Question

how to find lim of this ?

Answer 1

\[\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{6+...}}}}}}\]

Answer 2

Have you tried finding a recursive definition for the above sequence yet?

Answer 3

yes i tried but ,no result
i know it is converge
but i cant find an upper bound

Answer 4

what recursive formula have you found?

Answer 5

I am helpless, just mark to be back later. Sorry for this

Answer 6

I think this formula should work:
\[a_1=1,\\
a_{n}=\sqrt{n+a_{n+1}},~n\ge1\]

Answer 7

thank "SithsAndGiggles"
but it not work in infinity ,because the term of n
i tried it before
but thanks for your hint

Answer 8

can someone suggest the way , to find an upper bound for it ?

Answer 9

@SithsandGIggles

Shouldnt it say
a_n+1 = sqrt( n + a_n)

Answer 10

did you try to change to rational exponents?

Answer 11

ok this is a series, first let's write the partial sums

a1 = sqrt(1)
a2 = sqrt(1 + sqrt(2))

Answer 12

yes , i tried rational exponents , but no result
I know it is converge to about 1.75
from numerical solution ,
but I cant bring an analytic method

Answer 13

if i find an upper bound , it will be easy

Answer 14

looking at the partial sums we have
a1 = sqrt(1)
a2 = sqrt(1+ sqrt(2))
a3 = sqrt(1+ sqrt(2 + sqrt(3)))
...
an = sqrt(1 + sqrt(2 + sqrt(3 + ... sqrt(n)))))

Answer 15

you found an upper bound empirically?

Answer 16

yes it looks like 2 is an upper bound

Answer 17

how did you see 2 such as upper bound ?

Answer 18

using a calculator, i just kept increasing terms

Answer 19

i did partial sums
sqrt(1), sqrt(1+ sqrt(2)), sqrt(1+ sqrt(2 + sqrt(3))) , ...

Answer 20

and I noticed that the decimals are converging, I have 1.757 so far

Answer 21

ok
i see
but using calculator is not analytic!

Answer 22

right

Answer 23

so you basically want the limit of