Can anyone walk me through this problem?

Question

Apply integration to the expansion 1/1+x = _∑^∞_n=1 (-1)ⁿ xⁿ = 1-x² - x³ + ... to show that for -1< x< 1, ln (1+x) =∑^∞_n=1 ((-1)^n-1 xⁿ ) / n

Answer 1

if I say what is integral of dt/(1+t) for t=0 to x the answer would be ln(1+x)

I say 0 to x since formally lnx= integral z=1 to x of dz/z. Substitute z=t+1

and when z=1, t=0. that is ∫ dt/(1+t) from t=0 to x same as ∫ dt/t from t=1 to x

∫ dt/(1+t) from t=0 to x

is ln(1+x) + C

if I say what is integral of ∑(-1)ⁿx⁽ⁿ⁺¹⁾ where n= 0 to infinity, the answer would be

x-x²/2+x³/3-+...+C

I know 1/(1+x) = ∑(-1)ⁿx⁽ⁿ⁺¹⁾

take integral of both sides to see ln(1+x)=x-x²/2+x³/3-+...

now the matter of when does the series match up to (converge) to ln(1+x)

(graph x-x²/2+x³/3+x⁴/4 to see similarity between x=-1 to x=1

to ln(1+x)),, it is a little stunning

we know from ratio test the series converges |x|<1

it (the series) needs to be tested at x=1 and x=-1

at x=1, it an alternating series with the terms approaching zero (convergers)

[which helps us compute ln2=2-(2²/2)+(2³/3)-/+...]

at x=-1 the series is not alternating and does not converge.

Answer 2

1/(1+x) = 1 -x +x²- x³ + x⁴ -x⁵ +x⁶-⋅⋅⋅⋅

∫dx/(1+x) = ln (1+x) = ∫ [ 1 -x +x²- x³ + x⁴ -x⁵ +x⁶-⋅⋅⋅⋅] dx =

= ∫ 1dx - ∫ x dx + ∫ x² dx - ∫ x³ dx +∫ x⁴ dx - ∫x⁵ dx+ ∫ x⁶ dx-⋅⋅⋅

= x - x² / 2 + x³ /3 - x⁴ /4 + x⁵/5 - x⁶ /6 +⋅⋅⋅⋅

∑ _n=1[ (-1 )^n-1 x ⁿ ] / n

Hence ln (1 + x ) = ∑ _n=1[ (-1 )^n-1 x ⁿ ] / n

RADIUS OF CONVERGENCE

lim |(a_n+1)/a_n| = lim | n x / (n+1) | =| x | <1 that is -1 < x < 1