Question

What will be the value of

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

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

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 $a_1=\sqrt{42}\le\sqrt{49}=7$. Suppose $a_k\le 7$. Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$. So by induction, $a_n$ is bounded above by 7 for all $n$.

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$. It should be quite clear that $a_2>a_1$. Suppose $a_k>a_{k-1}$. Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$. 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,

$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L$

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

$\implies L=\sqrt{42+L}$

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.