Recent content by Entertainment Unit

Thanks for the help! I finally got ##\frac {28035698399}{158753389900} \approx 0.177##.

My answer is as follows: Let ##S## be the set of all outcomes of dealing four labelled 13-card hands from a standard 52-card deck. Let ##A## be event "N and E have exactly the same number of spades." Let ##A_i## be event "N and E have exactly ##i## spades each." Note that when ##i > 6##...

Thank you, the next section in my textbook is entitled "Power Series" so once I'm done this set of exercises, I'll work through that and come back to your post. Thanks again!

Let ##f(x) = \frac {\sqrt x} {e^\sqrt x}## Note f is a continuous, positive function on ##[1, \infty)## Note also that ##\frac {d} {dx} f(x) = \frac {1 - \sqrt x} {2e^{\sqrt x} \sqrt x} \leq 0## when ## x \geq 1 \implies## f is a decreasing function. Let ##u = e^{-\sqrt x} \implies du = \frac...

Point taken, I need to get over my laziness and just calculate the integral. I was hoping for one of those "why couldn't / didn't I think of that" moments in the form of a series to use to compare the given series with.

Homework Statement Test the following series for convergence or divergence. ##\sum_{n = 1}^{\infty} \frac {\sqrt n} {e^\sqrt n}## Homework Equations None that I'm aware of. The Attempt at a Solution I know I can use the Integral Test for this, but I was hoping for a simpler way.

Thanks, here's what wound up with: It is given that ##a_n = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)}## ##\implies a_{n + 1} = \frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot (3n - 2)(3n + 1)}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n + 1)(2n...

Homework Statement Determine whether the given series is absolutely convergent, conditionally convergent, or divergent. ##\frac{1}{3} + \frac{1 \cdot 4}{3 \cdot 5} + \frac{1 \cdot 4 \cdot 7}{3 \cdot 5 \cdot 7} + \frac{1 \cdot 4 \cdot 7 \cdot 10}{3 \cdot 5 \cdot 7 \cdot 9} + \ldots + \frac{1...

Yes, and by extension ##b_1 < b_2 < b_3 < \cdots < b_n < b_{n + 1} < a_{n + 1} < a_n < \cdots < a_3 < a_2 < a_1## which implies ##b_1 < a_n## and ##b_n < a_1##. It follows that, ##\{a_n\}## is bounded below by ##b_1 = \sqrt{ab}## and ##\{b_n\}## is bounded above by ##a_1 = \frac {a + b} 2##.

Ok, I think I see what you were getting at now. If ##\{a_n\}## were not bounded from below, it would run into ##\{b_n\}## at some point since ##a_n > a_{n+1} > b_{n+1} > b_n##. So, ##\{a_n\}## on the way down would run into ##\{b_n\}## on the way up as ##{n\to\infty}##. So the question...

I think you're getting at ##\lim_{n\to\infty} a_n## would equal ##-\infty## and to calculate this limit and see what actually happens?

Homework Statement Let ##a## and ##b## be positive numbers with ##a \gt b##. Let ##a_1## be their arithmetic mean and ##b_1## their geometric mean: ##a_1 = \frac {a + b} 2## and ##b_1 = \sqrt{ab}## Repeat this process so that, in general ##a_{n + 1} = \frac {a_n + b_n} 2## and ##b_{n + 1} =...

Thanks for this!

Thank you! I went down the path of proving it directly but struggled with that too. Here's my completed proof: Proof: Suppose $$\frac {b^{n + 1} - a^{n + 1}} {b - a} < (n+1)b^n$$ It follows that $$\frac {(b - a)(b^n + ab^{n - 1} + \dots + a^{n - 1}b + a^n)} {b - a} < (n+1)b^n$$ $$b^n + ab^{n...

Homework Statement Show that if ##0 \leq a < b##, then $$\frac {b^{n + 1} - a^{n + 1}} {b - a} < (n + 1)b^n$$ Homework Equations None that I'm aware of. The Attempt at a Solution Proof (Induction) 1. Basis Case: Suppose ##n = 1##. It follows that: $$\frac {b^{1 + 1} - a^{1 + 1}} {b - a} < (1 +...