Çözüldü Radius and Interval of Convergence In Series

https://i72.servimg.com/u/f72/19/97/10/39/radius10.png

u(n) = [ (x + 1)^n ] / n
u(n + 1) = [ (x + 1)^(n + 1) ] / (n + 1)
| u(n + 1) / u(n) | = | { [ (x + 1)^(n + 1) ] / (n + 1) } / { [ (x + 1)^n ] / n } |
| u(n + 1) / u(n) | = | [ n / (n + 1) ]·(x + 1) |
lim (n → ∞) = 1·| (x + 1) | < 1
|x + 1| < 1
-1 < x + 1 < 1
-2 < x < 0
For x = -2, the series becomes [ (-1)^(n + 1) ]·[ (-2 + 1)^n ] / n = -1 / n, the famous divergent harmonic series.
For x = 0, the series becomes [ (-1)^(n + 1) ] / n....(I)
Recalling ln(1 + x) = (lower limit n1 = 1, upper limit n2 = ∞),Σ [ (-1)^(n + 1) ]·(x^n) / n and substituting x = 1,
ln(1 + 1) = (lower limit n1 = 1, upper limit n2 = ∞),Σ [ (-1)^(n + 1) ]·(1^n) / n
ln2 = (lower limit n1 = 1, upper limit n2 = ∞),Σ [ (-1)^(n + 1) ] / n....(II)
Comparing (I) and (II), the series is concluded to be convergent for x = 0
That is; the interval of convergence is -2 < x ≤ 0, and the Radius of Convergence is R = 1
Choice C is correct.

WolframAlpha Check: https://www.wolframalpha.com/input/?i=sum (((-1)^(n+1))/n),n=1..infinity