1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
arlik [135]
3 years ago
6

Use the Integral Test to determine whether the series is convergent or divergent

Mathematics
1 answer:
Inga [223]3 years ago
7 0

Answer:

A. \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges by integral test

Step-by-step explanation:

A. At first we need to verify that the function which the series is related (\frac{n}{e^{15n}}) fills the necessary conditions to ensure that the test is effective.

*f(x) must be continuous or differentiable

*f(x) must be positive and decreasing

Let´s verify that f(x)=\frac{n}{e^{15n}} fills these conditions:

*Considering that eˣ≠0 for all x, the function f(x)=\frac{n}{e^{15n}} does not have any discontinuities, so it´s continuous

*Because eˣ is increasing:

      if a<b ,then eᵃ<eᵇ

      if 0<eᵃ<eᵇ ,then 1/eᵃ > 1/eᵇ

      if 1/eᵃ > 1/eᵇ and a<b, then a/eᵃ<b/eᵇ

  We conclude that f(x)=\frac{n}{e^{15n}} is decreasing

*Because eˣ is always positive and the sum is going from 1 to ∞, this show that f(x)=\frac{n}{e^{15n}} is positive in [1,∞).

Now we are able to use the integral test in f(x)=\frac{n}{e^{15n}} as follows:

\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges

Let´s proceed to integrate f(x) using integration by parts

\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx

Choose your U and dV like this:

U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}

And continue using the formula for integration by parts:

\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}

Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}}-(\frac{-1}{15e^{15}}-\frac{1}{225e^{15}})

\int_{1}^{\infty}xe^{-15x}\ dx= ( \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}})+\frac{1}{15e^{15}}(1-\frac{1}{15})

We only have left to solve the limits, but because b goes to  ∞ and it is in an exponential function on the denominator everything goes to 0

\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})

Showing that the integral converges, it´s the same as showing that the series converges.

By the integral test \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges

You might be interested in
What is the total cost to repay a credit card loan with a $13,000 balance, an APR of 17%, and monthly payments of $220?
iren2701 [21]
It will take you 130 months to pay off this credit card.  You will pay $15,411.19 in interest over this period.
3 0
3 years ago
3.4-2.8d+2.8d-1.3. Combine like terms to simplify the expression
tekilochka [14]
   
<span>3.4 - 2.8d + 2.8d - 1.3 = </span>3.4 - 1.3 - 2.8d + 2.8d = 3.4 <span>- 1.3 = 2,1



</span>
8 0
3 years ago
A soccer player ran 180.48 miles in 70.5 days to stay fit. How many miles will he run in 9 days
cluponka [151]

Answer:

23.04 miles

Step-by-step explanation:


4 0
3 years ago
Let the probability of success on a Bernoulli trial be 0.26. a. In five Bernoulli trials, what is the probability that there wil
andreyandreev [35.5K]

Answer:

0.3898 = 38.98% probability that there will be 4 failures

Step-by-step explanation:

A sequence of Bernoulli trials forms the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Let the probability of success on a Bernoulli trial be 0.26.

This means that p = 0.26

a. In five Bernoulli trials, what is the probability that there will be 4 failures?

Five trials means that n = 5

4 failures, so 1 success, and we have to find P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{5,1}.(0.26)^{1}.(0.74)^{4} = 0.3898

0.3898 = 38.98% probability that there will be 4 failures

4 0
3 years ago
What is 12 2/3 as an improper fraction
BaLLatris [955]

Answer:

\dfrac{38}{3}

Step-by-step explanation:

12\dfrac{2}{3} =

= 12 + \dfrac{2}{3}

= \dfrac{12}{1} + \dfrac{2}{3}

= \dfrac{12}{1} \times \dfrac{3}{3} + \dfrac{2}{3}

= \dfrac{36}{3} + \dfrac{2}{3}

= \dfrac{38}{3}

4 0
3 years ago
Other questions:
  • you think of a number from the first 20 negative integers what is the probability that the integers chosen will be divisible by
    14·1 answer
  • 0.0000003513 in scientific notation
    11·2 answers
  • Please help me with missing angles​
    7·1 answer
  • According to the Square Root property of Equations, the solution set of x2 = 25 is {±5}.
    8·1 answer
  • 1/2 = 1/4 (×-10) how can I solve it
    10·1 answer
  • Stainless steels are frequently used in chemical plants to handle corrosive fluids, however, these steels are especially suscept
    13·2 answers
  • I need help plz!!!!!
    9·2 answers
  • SOMEONE HELP !! Asap! I’ll give brainliest!!
    13·1 answer
  • How do i factor equations?
    12·1 answer
  • Evaluate cos 150° without using a calculator.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!