Author Topic: TT's Maths Thread (Read 176533 times) Tweet Share

TrueTears · « **Reply #900 on:** February 09, 2010, 03:36:43 pm »

thanks humph.

Also using the test for divergence on any alternating series, wouldn't the limit would never exist.

eg, $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1}$

$\lim_{n \to \infty} (-1)^n \frac{n}{n+1}$ doesn't exist so the series is diverging, but it is actually converging if we use the alternating series test?

kamil9876 · « **Reply #901 on:** February 09, 2010, 03:55:16 pm »

Quote from: humph on February 09, 2010, 02:55:19 pm

Quote from: TrueTears on February 09, 2010, 02:35:51 pm
thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

You want to show that $n! > 10^n$ for sufficiently large $n$ - work out when this inequality is first true, then use induction (this is when a calculator or computer comes in handy).

That's still too weak though.

for sufficiently large n the following is true:

$\frac{10^n}{n!}=\frac{10}{1} \times \frac{10}{2} ..... \times \frac{10}{10} \times \frac{10}{11} \times \frac{10}{12}.... \frac{10}{n} \leq \frac{10^{10}}{10!} \times (\frac{10}{11})^{n-10}$

So you can compare it to a geometric series with first term $\frac{10^{10}}{10!}$ and common ratio $\frac{10}{11}$

TrueTears · « **Reply #902 on:** February 09, 2010, 04:05:52 pm »

So you're saying we can't use the test for divergence on alternating series?

how come stewarts did it on page 712?

oh u deleted ur post lol

kamil9876 · « **Reply #903 on:** February 09, 2010, 10:18:57 pm »

Quote from: TrueTears on February 09, 2010, 03:35:33 am

If $\sum a_n$ is a conditionally convergent series and $r \in \mathbb{R}$ then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$ .

Argh, I haven't done a proof in ages, but this one really caught my attention xD

Let $(u_i)$ be the subsequence of $(a_i)$ that consists of positive terms and let $(l_i)$ be the subsequence of $(a_i)$ that consists of negative terms. Specifically, let:

$u_i=\Big\{^{a_i, a_i>0}_{0, otherwise}$

and likewise for $l_i$

Lemma: $\sum u_i$ and $\sum l_i$ are both diverging.

The following construction produces the rearangement for the given $r$ :

Let $(x_i)$ be this rearangment. Imagine that it is initially empty, we will now create it by adding in the numbers in order:

Keep adding the first few numbers from $(u_i)$ until the sum exceeds $r$ . Then keep adding the first few numbers from $l_i$ until the cumulutative sum drops below $r$ . Then keep repeating this for the next lot of numbers in $u_i$ and $l_i$ ad infinitum.

Now I will leave it for you to prove that such a construction can be done (you will need the lemma) and that the limit of (x_i) is indeed $r$ .

kamil9876 · « **Reply #904 on:** February 10, 2010, 12:34:47 pm »

Extension question: Prove that if $\sum_{i=1}^{\infty} a_{\phi(i)}$ is a rearangement of a convergent series $\sum_{i=1}^{\infty} a_i$ such that for some fixed integer $p$ , $|\phi(i)-i|\leq p$ (ie each term is "shifted" at most $p$ places) then the two sums are equal.

TrueTears · « **Reply #905 on:** February 10, 2010, 03:15:55 pm »

$\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right)$

diverging or converging?

First I thought integral test would work since $\tan\left(\frac{1}{n}\right)$ is decreasing for $n \ge 1$ and positive. But the integral is not elementary.

Using the divergence test we get the limit as $0$ which is inconclusive.

Next, I tried to use the ratio test.

$\lim_{n \to \infty} \left | \frac{\tan\left(\frac{1}{n+1}\right)}{\tan\left(\frac{1}{n}\right)} \right |$

Since both functions are positive for $n \ge 1$ , the modulus signs are redundant.

$\lim_{n \to \infty} \frac{\tan\left(\frac{1}{n+1}\right)}{\tan\left(\frac{1}{n}\right)} = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \cdot \lim_{n \to \infty} \frac{\cos\left(\frac{1}{n}\right)^2}{\cos\left(\frac{1}{n+1}\right)^2} = 1$

Which again is inconclusive.

So what do I do...?

EDIT: 6000th post yay!

kamil9876 · « **Reply #906 on:** February 10, 2010, 04:06:05 pm »

$lim_{n\to \infty} \frac{tan(\frac{1}{n})}{\frac{1}{n}}=lim_{x\to 0} \frac{tan(x)}{x}=\lim_{x\to 0}\frac{sin(x)}{x} * \lim_{x \to 0} \frac{1}{cos(x)}=1$

Hence either both the harmonic series and the series in question converge, or both diverges. Meaning the series in question diverges.

TrueTears · « **Reply #907 on:** February 10, 2010, 04:17:52 pm »

thanx

TrueTears · « **Reply #908 on:** February 10, 2010, 04:58:41 pm »

$\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$

Using the root test we get:

$\lim_{n \to \infty} \left | \left(\frac{(n!)^n}{n^{4n}}\right)^\frac{1}{n} \right | = \lim_{n \to \infty} \frac{n!}{n^4}$

So... how to evaluate that haha

kamil9876 · « **Reply #909 on:** February 10, 2010, 05:19:54 pm »

for $n>4$

$\frac{n!}{n^4}=\frac{n}{n} \times \frac{n-1}{n} \times \frac{n-2}{n} \times \frac{n-3}{n} \times (n-4)!$

$=(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n}) \times (n-4)!>(1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})(n-4)!$

Hence the limit is infinity.

TrueTears · « **Reply #910 on:** February 10, 2010, 05:48:03 pm »

Your inequality shouldn't it be

$(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n}) \times (n-4)! \ge (1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})(n-4)!$

coz they can be equal if n = 4, it's not undefined for n = 4...? why are you saying it is only greater?

kamil9876 · « **Reply #911 on:** February 10, 2010, 05:54:34 pm »

sure, you can do that if you want. Getting the maximal domain is not important, my one still does the job.

TrueTears · « **Reply #912 on:** February 10, 2010, 06:10:15 pm »

yeah, true, thanks

Mao · « **Reply #913 on:** February 10, 2010, 10:58:19 pm »

Just a note: have you looked at the ratio test?

Taking $\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$ for example

$|r_n| = \frac{10^n}{n!}$

$|r_{n+1}| = \frac{10^{n+1}}{(n+1)!}$

$\lim_{n\to \infty} \frac{|r_{n+1}|}{|r_{n}|} = \frac{10}{n+1} = 0$

:. Absolute convergence

It can make life a lot easier: http://en.wikipedia.org/wiki/Ratio_test

TrueTears · « **Reply #914 on:** February 10, 2010, 10:58:34 pm »

$\sum_{n=1}^{\infty} \left( \frac{n}{n+1} \right)^{n^2}$

So applying the root test we get:

$\lim_{n \to \infty} \left | \left( \frac{n}{n+1} \right)^n \right | = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n$

So let $y = \left( \frac{n}{n+1} \right)^n$

Thus $\log_e(y) = \log_e\left(\left( \frac{n}{n+1} \right)^n\right) = n \log_e\left( \frac{n}{n+1} \right)$

$\implies \lim_{n \to \infty} \log_e(y) = \lim_{n \to \infty} n \log_e\left( \frac{n}{n+1} \right)$

I've used hospital on $\lim_{n \to \infty} n \log_e\left( \frac{n}{n+1} \right)$ but it's too complicated...

Nvm fkn stupid arithmetic error grr

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