# how you solve: #lim_(n->oo)(2-sqrt2)(2-root(3)2)(2-root(4)2)...(2-root(n)2))# ?

The necessary and sufficient condition for convergence of

Considering now

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So:

Then:

So:

So:

diverges, since the harmonic series diverges.

So:

and therefore:

For reference, here's a failed attempt that nevertheless illustrates a way you might approach such a problem...

The basic idea is:

Hmmm...that does not quite work either, since:

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To solve the limit (\lim_{n \to \infty} (2 - \sqrt{2})(2 - \sqrt[3]{2})(2 - \sqrt[4]{2}) \ldots (2 - \sqrt[n]{2})), we recognize that each term in the product is of the form (2 - \sqrt[k]{2}), where (k) varies from (2) to (n).

Now, let's rewrite each term in a way that allows us to use a limit property. We can express (2 - \sqrt[k]{2}) as (\sqrt[k]{2} - 2). Then, we can rewrite the expression as (\lim_{n \to \infty} (\sqrt[k]{2} - 2)) as (k) ranges from (2) to (n).

Next, we notice that as (k) approaches infinity, (\sqrt[k]{2}) approaches (1). Therefore, (\sqrt[k]{2} - 2) approaches (1 - 2 = -1).

Since the limit of each term in the product approaches (-1), and we have an infinite number of such terms, the overall limit of the product is ((-1)^{\infty}).

However, ((-1)^{\infty}) is an indeterminate form. To resolve this, we can apply the natural logarithm and rewrite the expression as (\lim_{n \to \infty} \ln[(2 - \sqrt{2})(2 - \sqrt[3]{2})(2 - \sqrt[4]{2}) \ldots (2 - \sqrt[n]{2})]).

Using the property (\ln(ab) = \ln(a) + \ln(b)), we can rewrite the expression as (\lim_{n \to \infty} [\ln(2 - \sqrt{2}) + \ln(2 - \sqrt[3]{2}) + \ln(2 - \sqrt[4]{2}) + \ldots + \ln(2 - \sqrt[n]{2})]).

As (n) approaches infinity, each term in the sum approaches (\ln(2 - 1) = \ln(1) = 0).

Therefore, the overall limit is (0).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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