A string is 5.8 meters long.

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

How to solve this equation 15(6-g)

NikAS [45]

15(6−g)=(15)(6+−g)=(15)(6)+(15)(−g)=90−15g=−15g+90

7 0

3 years ago

Please help with this question thank you very much.

DochEvi [55]

Answer:

ln 3

Step-by-step explanation:

3 ln 3 - ln 9

We can rewrite 9 as 3^2

3 ln 3 - ln 3^2

We know lnx^a = aln x

3 ln 3 - 2 ln 3

ln 3

6 0

3 years ago

a weather man is found to be correct 80% of the time ( 80 out of 100 ) if he gives 375 forecasts how many of them will be correc

Galina-37 [17]

Bro don’t click the file or the link

3 0

3 years ago

A miner descended 1200 feet into a mine. While in the mine, he then rose 900 feet and then descended 350 feet. What was the mine

solniwko [45]

We can just add all these numbers together because they are just addition and subtraction.

We assume the miner starts at 0 (the surface).

0 - 1200 (down) + 900 (up) - 350 (down) = -650 (C)

6 0

3 years ago