Does the following infinite geometric series diverge or converge? Explain.

1 answer:

6 0

Your answer is Diverge. To tell whether a series is divergent or convergent, you must look at the rate (the number it is being multiplied by) for the series. If the rate is less than 1, then it would be convergent, meaning it does have a sum. If the rate is greater than 1, then that means it does not have a sum and never will. These series will diverge. Now your rate in this problem is 3 and 3 is greater than 1, so it diverges.

You might be interested in

A clothing store buys jeans for $12 and marks them up 200% before selling them to customers. What is the selling price of the je

Ganezh [65]

Answer:

$24

Step-by-step explanation:

200% of 12 is 24

5 0

2 years ago

Read 2 more answers

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given indefinite integral:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$