Intergal (arctan(x^2),x,0,3/4) aproximate the value using Rn <= 0.001

I totally can do this.. if it didn't have an intergal... I know arctan(x)=~ x-x^3/3+x^5/5-x^7/7+... + ((-1)^nx^(2n+1))/(2n+1)

thus arctan(x^2)=Sigma((-1)^n*(x^2)^(2n+1))/(…

but that integral...?

Oh also I cant find the convergence / divergence of

Sigma(((-1)^n*n)/Ln(n),n,1,infinity) that Ln is really mucking it up. My solution guide says Nth term test and skips the work ( its so helpful!) this one is not a big priority, but there must be a simple solution..

Hey so my teacher is really strict on these writing questions. Can you check my work?

For any elementary function y = f (x) , describe the process used to approximate f (a) to the nearest thousandth using the Maclaurin Series for the function where a is within the interval of convergence.

As its a Maclaurin Series C=0

I would write find the first 4 deriviatives. Then plug in Center = 0 for x.

f(x) to f(0)

f'(x) to f'(0)

f''(x) to f'''(0)

f'''(x) to f'''(0)

f^4(0) to f^4(0)

then plug in

Psub4 = f(0)/0!*(x-0)^0+f'(0)/1!*(x-0)^1+f''(0)/… then look for a pattern.

we have a sigma now. sigma(otherTerms*x^n) taking either a geometric, integral, or root test ( that IS correct Right? Those are the only tests I will try?) I'll get | x |< 1 converges | x | > 1 diverges. (or possibly | x+-k |< 1 converges | x+-k | > 1 diverges. but as this is a Maclorin series this is unlikely we will get)

lets assume we do get |x|=1 from a series test. which goes to x=+-1

-1<x<1 for our interval of convergence (or (-1,1))

now for error we plug in, well ANY number within the interval of convergence. I'll choose 0.5

back to sigma(otherTerms*x^n) which goes to sigma(otherTerms*(0.5)^n,n,0,infinity)

next plug in N terms until we find Rn>0.001 (Rn less then 1/1000)

Thanks for your help!