How can I check my Maclaurin series?

[HappyCalculusStudent]

I’m working on finding the first few terms of the Maclaurin series for arcsinh (x) - (inverse hyperbolic sin of x).
I want to do it several ways if I can, but one I have to do is by integrating its derivative: 1/sqrt(1+x)
So, upon integration I got x, -x^2/4, x^3/8, -5x^4/64, ….
I tried to see if this is right by squaring 1/ sqrt (1+x) to get the power series of 1/ 1+x which I’m familiar with, but I’m having trouble. I can’t think of how else to check my work.
Could someone let me know if this is right or wrong, and what’s the best way to check my answers with a problem like this one?

Wikipedia had this (see attachment):

but my series doesn’t seem to match this when I play with the form in summation (sigma) notation.


One more quick thing:
{ sin n / sqrt n }
My book says it goes to zero by the Squeeze Theorem.
I see that it is smaller than { 1/ sqrt n }, but do I have to use absolute value? Why is the lower bound zero?

 

[tkhunny]

You may wish to check that derivative. Are you sure it's 'x' in the denominator?

Ohm and good luck squaring that series. That is an activity we like to call "essentially impossible". It can be done, but why on Earth would you want to?

 

[BigGlenntheHeavy]

HappyCalculusStudent, if you have a trusty TI-89, plug in F3-9 and you'll get a Taylor series.

The first ten terms agree with your attachment, note, every even term goes to zero.

Viz., f(x) = arcsinh(x) = x -x^3/6+(3x^5)/40-(5x^7)/112+(35x^9)/1152-...+...-...+....

 

[BigGlenntheHeavy]

\(\displaystyle Find \ \lim_{x\to\infty} \frac{sin(x)}{\sqrt x}. \ Not \ much \ help \ from \ the \ Marquis \ on \ this \ one, \ but \ not \ to \ worry,\)
\(\displaystyle as \ "The \ Squeeze \ Theorem" \ to \ the \ rescue.\)

\(\displaystyle We \ know \ that \ sin(x) \ is \ greater \ than \ or \ equal \ to \ minus \ one \ and \ less \ than \ or \ equal \ to \ one.\)

-1 ? sin(x) ? 1. Dividing the inequality by ?x, we get -1/?x ? sin(x)/?x ? 1/?x.

\(\displaystyle Now \ \lim_{x\to\infty} \frac{-1}{\sqrt x} \ and \ \lim_{x\to\infty} \frac{1}{\sqrt x} \ both \ equal \ 0.\)

Hence, this implies that 0 ? sin(x)/?x ? 0, which implies that the limit of sin(x)/?x = 0 as x approaches infinity. QED.

 

[HappyCalculusStudent]

Oops! Yes, I was missing the x^2 in the denominator. Thank you for pointing that out!

I just redid my work quickly, and from my scribbles I see that the first three terms now do agree with both Wikipedia's and BigGlenn's.
I feel much better about it.
And, now I'm starting to see how useful the sqeeze theorem can be.

Thank you so much for your help!