# Yesterday's test

1. Nov 20, 2004

### twoflower

Hi, yesterday we wrote test from calculus

1. Problem
Find out, for which $y \in \mathbb{R}$ the sum

$$\sum_{n=2}^{\infty} a_{n}$$

converges, where

$$a_{n} = \frac{1}{2^{n} - 2^{-n}} \frac{y^{n}}{2n - 1}$$

and for which y this sum converges absolutely.

2. Problem
Decide, whether the limit

$$\lim_{n \rightarrow \infty} b_{n}$$

exists, where

$$b_1 = 4,$$

$$b_{n+1} = \frac{6}{1 + b_{n}} \forall n \in \mathbb{N}$$

If the limit exists, find it.

3. Problem
Decide, whether the limit exists:

$$\lim_{n \rightarrow \infty} \frac{3^{n} + 2n^{n} + n!}{(n+1)^4 + \sin{n} + (3n)!}$$

If the limit exists, find it.

-----------------------

Here's how I approached:
1. Problem
First I try interval $y \in <0,2)$ (I just guessed it...).
I used Abel's-Dirichlet's theorem about the convergence and I proved the preconditions so that I could say the sum converges for y in this interval.

For interval $y \in (-2, 0)$ I used Leibniz's theorem and I found out that it converges in this interval too.

In sum, we get that the sum converges for $y \in (-2, 2)$.

For every other y the sum diverges I think.

The sum converges absolutely for $y \in (-2, 2)$ too.

2. Problem
I was quite useless at solving this, I tried to prove the convergence using Bolzano-Cauchy criterion, but I got somewhat non-sense result (limit in $\mathbb{R}$ doesn't exist)

3. Problem
Next problem I wasn't completely sure about. I multiplied the limit with

$$\frac{\frac{1}{n^{n}}}{\frac{1}{n^{n}}}$$

and comparing the $n^{n}$ and $(3n)!$ (which gets larger in infinity I think) I got that the limit is 0. Is it ok?

2. Nov 20, 2004

### Inquisitive_Mind

Hint for Q2: Investigate the absolute difference $$|b_{n}-b_{n-1}|$$, noting that $$b_{n}>0$$ for all n. You should get something like $$|b_{n+1}-b_{n}|\leq 6|b_{n}-b_{n-1}|$$. Then use Cauchy's criterion for convergence.

3. Nov 21, 2004

### Astronuc

Staff Emeritus
4. Nov 21, 2004

### arildno

Astronuc:
That is wrong:
$$(3n)!=n!\prod_{i=1}^{n}(n+i)\prod_{j=1}^{n}(2n+j)\geq{n!}n^{2n}$$
Hence the limit exists, and equals 0

5. Nov 22, 2004

### devious_

Out of curiousity, what level calculus is this?

6. Nov 22, 2004

### twoflower

I don't know what exactly you mean with "level", but it was compulsory test in first semester. Of course it seems primitive for your eyes but we're beginners in fact :)

7. Nov 22, 2004

### twoflower

Beside the fact what you stated is not right according to me, we had to do it without l'Hospital.

We don't know differentiating yet...

8. Nov 22, 2004

### Astronuc

Staff Emeritus
I got my tail whipped on that one. My apologies for my mistake.

Certainly n^n blows up faster than n! (and 3*n!), but not as fast as(2n)! and certainly not (3n)!

And forget L'Hospital's rule.

Arildno, in what reference did you find that identity or did you derive it?

9. Nov 23, 2004

### twoflower

Arildno, are you saying that the whole original limit is 0, or that just

$$\frac{n^{n}}{(3n)!}$$

goes to 0?

Because Maple gives me "undefined" when I give him the whole limit...

10. Nov 23, 2004

### Dark Knight

i had an exam yesterday too:

find the nth derivative for:

$$f(x) = x^3 cos^2x$$

where n>3

can sombody help me?!!!

11. Nov 23, 2004

### arildno

Rewrite:
$$\cos^{2}x=\frac{1+\cos(2x)}{2}$$
Hence,
$$f(x)=\frac{x^{3}}{2}+\frac{x^{3}}{2}\cos(2x)$$
(Now you see why they have chosen n>3..)

In order to find the n'th derivative of the product, verify that if h(x)=f(x)g(x),
then the n'th derivative of h satisfies:
$$h^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}f^{(n-k)}(x)g^{(k)}(x)$$
that is, use of the binomial formula.

Then set
$$g(x)=\frac{x^{3}}{2}, f(x)=\cos(2x)$$
The summation then stops at k=3.

12. Nov 23, 2004

### arildno

The whole limit goes to zero.
To explain Maple's response, there are two options:
1) You haven't typed it correctly.
If for example, you typed 3*n! rather than (3*n)!, the limit would not exist.
(I haven't used Maple myself, so I don't know which typo's would most likely occur)
2) Maple has tried to calculate ratios of numbers too big to handle.
I'm rather skeptical of this option, since Maple is a professional software program which ought not be sensitive to this type of problems.
However, it depends on how Maple actually calculates its values, and I don't know how it does that..

13. Nov 23, 2004

### twoflower

Now I tried it again and I may really have typed it wrong, because now Maple says it doesn't know how to do the limit, which is much more logical answer.

14. Nov 23, 2004

### arildno

Note that both the numerator and denominator grows exceptionally fast; you will for rather small n's reach huge numbers.

I would suggest you type in for a few values of n (say, up to 10 or so); the decreasing pattern should be obvious.