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Here are some general hints on solving this question:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function  $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't needed for the solution; it is included for completeness.) Using a well-known rule for the expected value of a non-negative random variable, we have: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.

Some further insights:

From the above working we see that this distributional result and resulting expected value do not depend on the underlying distribution, so long as it is a continuous distribution. This is really not surprising once we consider the fact that every continuous scalar random variable can be obtained via a monotonic transformation of a uniform random variable (with the transformation being its quantile function). Since monotonic transformations preserve rank-order, looking at the probabilities of orderings of arbitrary IID continuous random variables is the same as looking at the probabilities of orderings of IID uniform random variables.

Here are some general hints on solving this question:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. The event $N \geqslant n$ means that there are at least $n$ increasing values at the start of the sequence, which is equivalent to saying that the first $n$ values are increasing. We therefore have the event equivalence:$$N \geqslant n \quad \quad \iff \quad \quad U_1 < U_2 < \cdots < U_n.$$Now, because the latter are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely. There are $n!$ possible orderings of the first $n$ values, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function:$$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't needed for the solution; it is included for completeness.) Using a well-known rule for the expected value of a non-negative random variable, we have: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.

Some further insights:

From the above working we see that this distributional result and resulting expected value do not depend on the underlying distribution, so long as it is a continuous distribution. This is really not surprising once we consider the fact that every continuous scalar random variable can be obtained via a monotonic transformation of a uniform random variable (with the transformation being its quantile function). Since monotonic transformations preserve rank-order, looking at the probabilities of orderings of arbitrary IID continuous random variables is the same as looking at the probabilities of orderings of IID uniform random variables.

Here are some general hints on solving this question:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't needed for the solution; it is included for completeness.) Using a well-known rule for the expected value of a non-negative random variable, we have: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.

Here are some general hints on solving this question:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't needed for the solution; it is included for completeness.) Using a well-known rule for the expected value of a non-negative random variable, we have: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.

Some further insights:

From the above working we see that this distributional result and resulting expected value do not depend on the underlying distribution, so long as it is a continuous distribution. This is really not surprising once we consider the fact that every continuous scalar random variable can be obtained via a monotonic transformation of a uniform random variable (with the transformation being its quantile function). Since monotonic transformations preserve rank-order, looking at the probabilities of orderings of arbitrary IID continuous random variables is the same as looking at the probabilities of orderings of IID uniform random variables.

UPDATE: I initially gave a full solution to this question, but I just realised that this might be a homework problem. In case it is a homework problem, here are some general hints on solving it:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't actually needed for the solution, but it is included for completeness.) The expected value of $N$ is: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.

Here are some general hints on solving this question:

You have a sequence of continuous IID random variables which means they are exchangeable. What does this imply about the probability of getting a particular order for the first $n$ values? Based on this, what is the probability of getting an increasing order for the first $n$ values? It is possible to figure this out without integrating over the distribution of the underlying random variables. If you do this well, you will be able to derive an answer without any assumption of a uniform distribution - i.e., you get an answer that applies for any exchangeable sequences of continuous random variables.

Here is the full solution (don't look if you are supposed to figure this out yourself):

Let $U_1, U_2, U_3, \cdots \sim \text{IID Continuous Dist}$ be your sequence of independent continuous random variables, and let $N \equiv \max \{ n \in \mathbb{N} | U_1 < U_2 < \cdots < U_n \}$ be the number of increasing elements at the start of the sequence. Because these are continuous exchangeable random variables, they are almost surely unequal to each other, and any ordering is equally likely, so we have: $$\mathbb{P}(N \geqslant n) = \mathbb{P}(U_1 < U_2 < \cdots < U_n) = \frac{1}{n!}.$$ (Note that this result holds for any IID sequence of continuous random variables; they don't have to have a uniform distribution.) So the random variable $N$ has probability mass function $$p_N(n) = \mathbb{P}(N=n) = \frac{1}{n!} - \frac{1}{(n+1)!} = \frac{n}{(n+1)!}.$$ You will notice that this result accords with the values you have calculated using integration over the underlying values. (This part isn't needed for the solution; it is included for completeness.) Using a well-known rule for the expected value of a non-negative random variable, we have: $$\mathbb{E}(N) = \sum_{n=1}^\infty \mathbb{P}(N \geqslant n) = \sum_{n=1}^\infty \frac{1}{n!} = e - 1 = 1.718282.$$ Note again that there is nothing in our working that used the underlying uniform distribution. Hence, this is a general result that applies to any exchangeable sequence of continuous random variables.