help with this calc problem plz?

Question

Find a). the radius of convergence for the series and b). it's interval of convergence. Then identify he values of x for which the series converges c). absolutely and d). conditionally

1). ∞
∑ (-x)^(n) / (n!)
n=0


∞
2). ∑ ((n+1)(2x +1)^n) / ((2n+1)(2^n))
n=0

Answer 1

1). A_n+1 / A_n = (-x) / (n+1) -> 0 as n-> ∞. So this series converges absolutely for all x (as we'd expect, recognising this to be the expansion of e^(-x).)

2). A_n+1 / A_n = (n+2)/(n+1) . (2x+1) . (2n+1)/(2n+2) . (1/2)
-> x + 1/2 as n -> ∞. So the radius of convergence is 1 and the interval of convergence has endpoints -3/2 and 1/2.
For x = -3/2, 2x+1 = -2 and we have
Σ(n=0 to ∞) (-1)^n (n+1) / (2n+1)
This is an alternating sequence with |An| -> 1/2, so is divergent.
For x = 1/2, 2x+1 = 2 and we have the same as for x = -3/2 but without the (-1)^n factor. This sequence is also divergent.
So the interval of convergence is (-3/2, 1/2); the series converges absolutely for x in this interval, and diverges outside it. There is no value of x for which this series is conditionally convergent.

Answer 2

i kinda dont have a clue but
1)compare to a greater series such as (2^n)/(n!)
2) whats inside the parenthesis can be written as the multiplication of the sums of two separate functions so
[(n+1)/(2n+1)]X[{(2x+1)^n}/(2^n)]

i think that should help, if not, then my bad

Answer 3

i will do quantity 3. If it develop into differentiable at x=0, the spinoff at the two factors could be equivalent. think of of honestly the fee function as a graph of y=-x whilst x is below 0 and a graph of y=x whilst x is larger than 0. The spinoff at y=-x is y'=-a million, however the spinoff at y=x is y'=a million, in spite of the element. consequently, you won't have the ability to have a spinoff at x=0. you need to continuously graph it and see that it does no longer artwork. i'm able to attempt a million. for any quantity ok, lim ok*f(x) = ok*lim f(x) x->c lim f(x) x->c = [f(x) - f(c)] / (x-c). lim ok f(x) x->c = [kf(x) - kf(c)]/(x-c) =ok [f(x) - f(c)] / (x-c) = ok*lim f(x) x->c no longer one hundred% particular if this could be a adequate evidence, in spite of the undeniable fact that it is going to help. 2? to tutor that the function is non-quit at x=0, you need to tutor that the decrease on the superb and the decrease on the left are the two equivalent to 0. lim f(x)x->0+ = 0 (logically, all of us comprehend that, however sin1/x may well be, x is drawing close 0) lim f(x)x->0- = 0 (comparable reason as above. overlook approximately sin a million/x. x is drawing close 0, and so is the function) lim f(x)x->0+ = lim f(x)x->0- =0. So we proved the 1st section. 2nd section: the by-manufactured from xsin1/x... by-manufactured from a million/x is -2/x^2 , the by-manufactured from sin is cosin, so the by-manufactured from sin a million/x is -2cosin(a million/x)/(x^2) (chain rule). Product rule: by-manufactured from xsin1/x is 1sin1/x -2xcosin(a million/x)/x^2= sin(a million/x) - 2 cosin(a million/x) / (x^2) EEK. properly, the spinoff would not exist at x=0 by fact the denominator of any term can on no account be 0. i think you need to tutor that utilising the decrease definition of a spinoff besides. lim f(x) x->a [f(x) - f(a)] / (x-a) permit a=0 and lim f(x) x->0 [f(x) - f(0)] / (x-0). We stated that at x=0, f(x) = 0 so the above simplifies to lim f(x) x->0 [f(x)]/ (x)= lim f(x) x->0 xsin(a million/x)/ (x)= lim f(x) x->0 sin(a million/x) it rather is undefined.