Author Topic: TT's Maths Thread (Read 157812 times) Tweet Share

TrueTears · « **Reply #870 on:** February 08, 2010, 02:25:08 am »

last one for the night:

$\sum_{n=1}^{\infty} \frac{1+n+n^2}{\sqrt{1+n^2+n^6}}$

Is this series converging or diverging?

Mao · « **Reply #871 on:** February 08, 2010, 04:28:57 am »

A second degree over a third degree --> probably divergent.

$\sum_{n=1}^{\infty}\frac{1 + n + n^2}{\sqrt{1 + n^2 + n^6}} = \sum_{n=1}^{\infty}\frac{n}{\sqrt{1 + n^2 + n^6}} + \sum_{n=1}^{\infty}\frac{1+n^2}{\sqrt{1 + n^2 + n^6}}$

Convergence $\sum_{n=1}^{\infty}\frac{n}{\sqrt{1 + n^2 + n^6}}$ is trivial.

For $n\in \mathbb{Z}^+$ :
$\frac{1+n^2}{\sqrt{1 + n^2 + n^6}} \ge \frac{1 + n^2}{\sqrt{1 + n^2 + n^6 + 2n^2 + 3n^4}}=\frac{1}{\sqrt{1+n^2}}$

$\frac{1}{\sqrt{1+n^2}} \ge \frac{1}{\sqrt{1 + n^2 + 2n}} = \frac{1}{n+1}$

$\implies \sum_{n=1}^{\infty}\frac{1+n^2}{\sqrt{1 + n^2 + n^6}} > \sum_{n=1}^{\infty} \frac{1}{n+1}$ , which is divergent.

Hence the series is divergent.

kamil9876 · « **Reply #872 on:** February 08, 2010, 12:29:04 pm »

Quote

A second degree over a third degree --> probably divergent.

That's a good intuition to have for solving this. Consequently it suggests that the term of highest degree is most significant, ie:

$\frac{1+n+n^2}{\sqrt{1+n^2 + n^6}}\geq \frac{n^2}{\sqrt{1+n^2 + n^6}}=\frac{1}{\sqrt{n^{-4} + n^{-2} + n^2}}$

$\geq \frac{1}{\sqrt{4n^2}}=\frac{1}{2n}$ Hence divergence.

TrueTears · « **Reply #873 on:** February 08, 2010, 03:50:49 pm »

Thanks u guys

TrueTears · « **Reply #874 on:** February 08, 2010, 04:49:52 pm »

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

Hmm what to compare this to? LOl

kamil9876 · « **Reply #875 on:** February 08, 2010, 04:52:57 pm »

hint:

$lim_{x\to0} \frac{sinx}{x}=1$

TrueTears · « **Reply #876 on:** February 08, 2010, 04:53:54 pm »

haha I see... limit comparison test!

Christiano · « **Reply #877 on:** February 08, 2010, 10:14:40 pm »

Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to.

O blessed math angels, come to my rescue.

the.watchman · « **Reply #878 on:** February 08, 2010, 10:15:54 pm »

$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread

TrueTears · « **Reply #879 on:** February 08, 2010, 10:16:23 pm »

TrueTears · « **Reply #880 on:** February 08, 2010, 10:16:41 pm »

Quote from: the.watchman on February 08, 2010, 10:15:54 pm

$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread

na it's ok, he can post here.

superflya · « **Reply #881 on:** February 08, 2010, 10:16:56 pm »

Quote from: Christiano on February 08, 2010, 10:14:40 pm

Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to. O blessed math angels, come to my rescue.

TT shall freak when he sees this

Edit: or maybe not.

the.watchman · « **Reply #882 on:** February 08, 2010, 10:20:37 pm »

Quote from: superflya on February 08, 2010, 10:16:56 pm

Quote from: Christiano on February 08, 2010, 10:14:40 pm
Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to. O blessed math angels, come to my rescue.

TT shall freak when he sees this

Edit: or maybe not.

LOL, don't go there superflya

TrueTears · « **Reply #883 on:** February 08, 2010, 10:22:26 pm »

superflya u pr0 kent

the.watchman · « **Reply #884 on:** February 08, 2010, 10:31:14 pm »

Quote from: TrueTears on February 08, 2010, 10:16:41 pm

Quote from: the.watchman on February 08, 2010, 10:15:54 pm
$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread
na it's ok, he can post here.

Soz

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