User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 February 19th, 2007, 07:01 AM #1 Senior Member   Joined: Jan 2007 From: India Posts: 161 Thanks: 0 To find the value of an infinite series friends , tel me if it is possible to compute the value of the below series without a calculator/computer. 1 + sqrt(2 + sqrt(3 + sqrt(4 + .........infinite terms ...) (Note:this series s not diverging .. it is in fact converging) February 19th, 2007, 07:49 AM   #2
Senior Member

Joined: Dec 2006

Posts: 1,111
Thanks: 0

I can't see how we would find what this series converged to, since it has no closed form. Any ideas? ( Actually, I think this is a sequence, not a series. )

Quote:
 Hope is a good thing. It might be the best of things. Nothing bad has ever come because of hope!
Unless, of course, you are hoping for something bad to happen.  February 19th, 2007, 08:51 AM #3 Senior Member   Joined: Jan 2007 From: India Posts: 161 Thanks: 0 hi infinity, actually i don understand the difference wen u mean as sequence and series ? anyway, for example say to find the value of the series , sqrt(12 + sqrt (12 + sqrt (12 ......to infinite terms )... this series might seem to diverge ..but not so actually. LET ,sqrt(12 + sqrt (12 + sqrt (12 ......to infinite terms )... = X sqrt(12 + X ) = X therfore, 12 + X = X^2 solvin X = 4 ...... ( u can chek this wit an ordinary calculator for the first few terms itself that it converges to 4) ... but the prob. is that the above method cant b used for the series in our case (but stil usin a calculator, the series (1 + sqrt(2 + sqrt(3 +.....) is seen to converge to 1.75 for first 10 terms) .... CAN this b got manually by solvin ??? February 19th, 2007, 09:12 AM #4 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 I'm not sure where you are getting 1.75, because that is less than 1+√2. I took the equation out to about 40 terms and came up with a value of 3.0903... February 19th, 2007, 10:22 AM   #5
Senior Member

Joined: Jan 2007
From: India

Posts: 161
Thanks: 0

Quote:
 Originally Posted by roadnottaken I'm not sure where you are getting 1.75, because that is less than 1+√2. I took the equation out to about 40 terms and came up with a value of 3.0903...

lets keep the series to b of finite terms (for convinience ) say as ,

1 + sqrt(2 + sqrt(3 + sqrt(4 + sqrt(5 + sqrt(6 + sqrt(7 + sqrt(8 + sqrt(9 + sqrt(10) .

comin from reverse.....

>> sqrt 10 = 3.162
>> 9 + 3.162 = 12.162
>>sqrt 12.162 = 3.487
>>8 + 3.487 = 11.487
>>sqrt 11.487 = 3.389
>>7 + 3.389 = 10.389
>>sqrt 10.389 = 3.223
>>6 + 3.223 = 9.223
>>sqrt 9.223 = 3.037
>>5 + 3.037 = 8.037
>>sqrt 8.037 = 2.834
>>4 + 2.834 = 6.834
>>sqrt 6.834 = 2.614
>>3 + 2.614 = 5.614
>> sqrt 5.614 =2.369
>> 2 + 2.369 = 4.369
>>sqrt 4.369 = 2.09
>> 1 + 2.09 = 3.09
>> sqrt 3.09 = 1.758 ...... hope u get it now .... February 19th, 2007, 02:41 PM #6 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 A sequence would be something like: n³/e^n from n=1 to n=∞ A series is the sum of all the terms of a sequence, and would be written as: ∑(n=1, n=∞) n³/e^n I assume that you are trying to find: Lim n→∞ { n + √(n+1 + √(N+2 + √(N+3 + ..... + √(∞)))) } Am I correct? February 19th, 2007, 02:46 PM   #7
Global Moderator

Joined: Nov 2006
From: UTC -5

Posts: 16,046
Thanks: 938

Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
 Originally Posted by Infinity I assume that you are trying to find: Lim n→∞ { n + √(n+1 + √(N+2 + √(N+3 + ..... + √(∞)))) } Am I correct?
That looks confusing and not well-defined (the square root of infinity?). Try this:

f(n) = n + sqrt(f(n+1))

Then arun is looking for f(1). February 19th, 2007, 03:00 PM #8 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Well, I just wrote this program to estimate the value of the sequence: Lim n→∞ { n + √(n+1 + √(N+2 + √(N+3 + ..... + √(∞)))) } to n = 100,000,000 The answer was: 1.7579327566180045 Code: import java.util.*; public class Sequence_Estimation { public static void main (String[] args) { long n=0; double answer=0; Scanner kbd = new Scanner(System.in); System.out.println("Input integer value of 'n' to test sequence to:"); n = kbd.nextLong(); for(long i=n; i>0; i--) { answer = Math.sqrt(answer + i); } System.out.println("\n" + answer); } } February 19th, 2007, 03:18 PM #9 Senior Member   Joined: Dec 2006 Posts: 1,111 Thanks: 0 Hey, I just calculated a more accurate answer to n = 10 billion, and guess what? It's exactly the same as the one I reported earlier. Apparently, more decimal places would be needed to show any change. February 19th, 2007, 03:56 PM #10 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms The number is called the nested radical constant. Its value is 1.757932756618004532708819638218138527653199922146 8377043 10135500385110232674446757572344554000259452970932 4718478 26956725286405867741108546115435116745974827649802 3843694 89120411842037876481995830644570345768467313417541 5134495 77173273720962022100603227554116598015407552297612 9445796 99112707719478877860007819516309923396999343623052 7753524 96605485188121304121230743966852549640366715265942 2159475 76652412589521440394432605735991324822082490634153 1503978 75302128772604959532494672112007991822456833844067 2864330 74237282346571947808094291349553420592279925860366 1703728 59630816687183328634908728532926587173888717587225 6906069 66741535388517308782986073313679762614334220034550 1474822 19697344628499290204994260780123338419145972718423 7910867 59045639529537528043251120937807502935923611917615 2704264 36487465911939829459953781691083134966345861642367 6784668 18801916873226676954205133566864879409563789163447 6743892 55347895570972640620596122532631802815634393718529 8175824 44581463125494708586493852134993196476027405424112 2516325 98737556657076790516333930301963846032409179377260 1377249 48433124123721498603941391880712274921521093576064 2271839 64712879727605419662075877641516168770731031830438 8844076... Tags find, infinite, series ,

,

,

,

,

,

,

,

,

,

,

,

,

,

# sqrt(12 sqrt(12 sqrt(12 ...... Infinity)=

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post math221 Calculus 2 April 6th, 2013 05:50 PM oddlogic Calculus 7 April 1st, 2011 04:49 PM oddlogic Calculus 3 March 31st, 2011 06:46 PM Zeefinity Real Analysis 3 April 22nd, 2010 09:58 PM johnny Calculus 1 September 9th, 2009 05:57 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      