Lim() will be approaching to 0.
Now, let’s find out the exact value for this infinity surd, which also could be known as Limit. We can work out this value by using the general formula found above:
In order to work it out in a easily, we need to get rid of the square root, there for square on both sides. That will give us .
And put “lim” signs on both side,
And using the value that we have found above, when* i) -> “k”
ii) -> “k”
K = K+1
K -K -1 = 0
And we can solve this equation by using quadratic function.
K = 1 ± (-1)2–4(1)(-1)
2(1)
K = 1 ± 5
2
K ≈ 1.618033989
Now, there is a positive value and negative value, which one is right..? the positive one will be the most logical answer. Because all of the values in the graph above are above zero, And the values in square roots were positive. Obviously, “K” value is positive.
We now have worked out the exact value(In other words, the Limit), therefore we know that the number will be keep increasing so close to the number, but it will never reach it.
Consider another infinity surd
The first term is given, .
The exact pattern that we found above can be observed here, therefore the general formula is very similar.
Use the same method, the first 10 series of values and the graph is:
b1 ≈ 1.8477590650226
b2 ≈ 1.9615705608065
b3 ≈ 1.9903694533444
b4 ≈ 1.9975909124103
b5 ≈ 1.9993976373924
b6 ≈ 1.9998494036783
b7 ≈ 1.9999623505652
b8 ≈ 1.9999905876192
b9 ≈ 1.9999976469034
b10≈ 1.9999994117258
Relationship between “n” and as “n” gets larger
We can see from this graph, we have got the similar results as above. The same convergence is happening here as well. From there is an rapid increase between the 1st value and 3rd value. After that, the value starts to increase only small numbers. We barely can see any changes on the graph, but we can know there is still a increasing number by the values we have calculated above.
Because there is a same pattern with previous one, this also suggests us that will also be approaching to zero.
We can use the same method we have used above to calculate the exact value(In other words, the Limit). I will simplified the formula and we get
B-B-2=0
We can solve this equation with same way with above, using quadratic function.
B = 1 ± (-1)2–4(1)(-2)
2(1)
B = 1 ± 9
2
Same reason, the data is also all positive, the most logical answer will be,
B = 2.
We now know the limit in this infinity surd is 2, so every numbers are trying to reach 2, but the numbers are going to be extremely close to 2,but will never reach it.
Now consider the general infinity surd, , where the first term is
By looking at the general formulas above, the first number in the root matched with the number in the Infinity surd, and after that, they had the same pattern of adding numbers behind. Therefore we can easily work out the general formula for this infinity surd. This is,
Xn+1 = k + Xn
We can calculate the exact value of this general infinity surd using the same method above, which it will give us
X-X-K=0
a = 1 ± (-1)2–4(1)(-k)
2(1)
a = 1 ± 1+4k
2
The pattern dictates that the answer must be positive, unless the value of “k” is less then zero. But it’s impossible in this case because negative number in a square root is not a real number.
The value of this infinity surd is not always an integer, that means some numbers make the infinity surd an integer. In order to be an integer, they can’t be decimal numbers, or fractions. So the numerator must be a divisor of the denominator to get integer numbers. In order to do that, the numerator must be a divisor of 2.
We have to get rid of square root sign if we want to get an integer. So the “k” value has to be a number which will lead numbers in the square root come out as an integer. And this number must be a divisor of 2 after adding up 1 to it.
But later I found out, it wasn’t necessary to see the number was divisor of 2 or not after adding 1 to it. Because when an integer came out of the square root, they were always odd number. And if you add 1 to it, it will become even number, therefore the number will always become divisor of 2 at all the time
Anyway, I have substituted all the positive numbers into K and I figured out which “k” values make the expression an integer. They are, 2,6,12,20,30, and so on. I thought zero should be included with one of them, but zero is just zero, it won’t have any increase of numbers in the infinity surd. Between in these numbers, 2,6,12,20,30, … … we can see there is a difference sequence. As you can see the difference between andis 4, and difference between and is 6, and the difference between and is 8, and so on. See the numbers of difference, 4,6,8, …… the numbers are constantly increasing by 2. This is telling us we can use “Difference integer” formula to find the general formula that represents all the values of “k” for which the expression is an integer.
The formula for Difference integer is,
The “n” has to be always greater or equal to 2 because “n” cannot be equal to zero, due to the fact that there is no such thing as.
Okay, we have 2,6,12,20,30, … … value for “k”,(the numbers which will always make the expression an integer)
And 4,6,8,10, … … for Difference(which is “” in this case)..
Therefore = 2.
And
Because the difference are the multiple numbers of 2
Therefore, = 2n.
Now let’s substitute the number’s in, 2 +
= 2 +
= 2 + (n-1)n
= n-n +2
Here we go, now we have worked out the general statement that represents all the values of “k” for which the expression is an integer.
Final conclusion is, “k” = n -n +2
