Question

What will be the value of


`\sqrt{42 + \sqrt{42 + √(42 + ....... \infty ) } }`
I don't really want the answer but the way to solve it.

Answer 1

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

`\begin{cases}a_1=√(42)\\a_n=\sqrt{42+a_(n-1)}\end{cases}`

Then you want to know the value of

`\displaystyle\lim_(n\to\infty)a_n`

if it exists.

To show the limit exists and that
`a_n` converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that
`a_1=√(42)\le√(49)=7`. Suppose
`a_k\le 7`. Then
`a_(k+1)=√(42+a_k)\le√(42+7)=7`. So by induction,
`a_n` is bounded above by 7 for all
`n`.

Monontonicity: We have
`a_1=√(42)` and
`a_2=\sqrt{42+√(42)}`. It should be quite clear that
`a_2>a_1`. Suppose
`a_k>a_(k-1)`. Then
`a_(k+1)=√(42+a_k)>\sqrt{42+a_(k-1)}=a_k`. So by induction,
`a_n` is monotonically increasing.

Then because
`a_n` is bounded above and strictly increasing, the limit exists. Call it
`L`. Now,

`\displaystyle\lim_(n\to\infty)a_n=\lim_(n\to\infty)a_(n-1)=L`

`\displaystyle\lim_(n\to\infty)a_n=\lim_(n\to\infty)\sqrt{42+a_(n-1)}=\sqrt{42+\lim_(n\to\infty)a_(n-1)}`

`\implies L=√(42+L)`

Solve for
`L`:

`L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7`

We omit
`L=-6` because our analysis above showed that
`L` must be positive.

So the value of the infinitely nested radical is 7.