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konstantin123 [22]
3 years ago
9

How to find the limit

Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0
\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}

By the Stolz-Cesaro theorem, this limit exists if

\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}

also exists, and the limits would be equal. The theorem requires that b_n be strictly monotone and divergent, which is the case since k\in\mathbb N.

You have

a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}

so we're left with computing

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}

This can be done with the help of Stirling's approximation, which says that for large n, n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n. By this reasoning our limit is

\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}

Let's examine this limit in parts. First,

\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}

As n\to\infty, this term approaches 1.

Next,

\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}

The term on the right approaches e^k, cancelling the e^{-k}. So we're left with

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}

Divide through the numerator and denominator by n^k:

\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}

So you can see that, by comparison, we have

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}

so this is the value of the limit.
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You roll a number cube twice. Find the probability of the events. Rolling an even number of 5.
lys-0071 [83]

Answer:

An even number and a 5. In the first roll you can get 2, 4 or 6, so you have 3 out of 6 options, the probability is 3/6 = 1/2, in the second roll you need you get a 5, the probability is 1/6.

I really hope this helps please give brainliest!!!!

5 0
3 years ago
What's the answer for 3/4 (x+3)=9
Vitek1552 [10]
We want to solve the equation :

\displaystyle{  \frac{3}{4}(x+3)=9


first, multiply both sides by 4:

\displaystyle{ \cancel{4} \cdot \frac{3}{ \cancel 4 }(x+3)=4\cdot 9

\displaystyle{ 3(x+3)=36


divide both sides by 3:

x+3=12


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Answer: x=9
3 0
3 years ago
Ralph is painting the barn below, including the sides and roof.  He wants to know how much paint to purchase. 
grandymaker [24]
PART A

The barn is constructed by the following 2D shapes

Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80

Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600

Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350

One base of length = 45' and width = 15'
Area = 45 × 15 = 675

Two rectangles on each on one side of the roof. We have the length = 45'  but not the width. We can work out the width by using Pythagoras theorem

w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77

Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33

Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)


PART B

Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans

PART C

Total cost of paint = 65 × 23.50 = $1527.50

PART D

The barn is constructed by a cuboid and a rectangular prism

V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500

V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800

Total volume = 1800 + 13500 = 15300


7 0
3 years ago
Solve:
riadik2000 [5.3K]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \dfrac{2x - 1}{5}  =  \dfrac{x -  2}{2}

\qquad \sf  \dashrightarrow \: 2(2x - 1) = 5(x - 2)

\qquad \sf  \dashrightarrow \: 4x - 2 = 5x - 10

\qquad \sf  \dashrightarrow \: 5x - 4x = -2 + 10

\qquad \sf  \dashrightarrow \: x = 8

Value of x is 8

8 0
2 years ago
Read 2 more answers
1). (-6p - 8) - (2p - 6) =
Alexeev081 [22]

Answer:

1.-8p-2

2.−4r−5

3.−7z−3

4.-3s-3

5. 1s

I got lazy to do the rest sorry :D

4 0
2 years ago
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