Find the radius of convergence and interval of convergence. ∑ n = 2 ∞ ( x + 2 ) n 2 n ln n ?

Question

Find the radius of convergence and interval of convergence.                     ∑                  n          =          2                ∞                                                  (            x            +            2            )                    n                                      2          n                          ln          n                    ?

Gustavo Navarro · Accepted Answer

Thus, lifting off from where you ended, let&#039;s apply the Ratio Test:            a              n        +        1              =                            (          x          +          2          )                          n          +          1                                      2                      n            +            1                                    ln                      (            n            +            1            )                                          a      n        =                            (          x          +          2          )                n                              2          n                          ln          n                    Therefore,      L    =          lim              n        →        ∞                    |                                    (            x            +            2            )                                n            +            1                                                2                          n              +              1                                            ln                          (              n              +              1              )                                          ⋅                                    2            n                                ln            n                                                (            x            +            2            )                    n                    |                                    (          x          +          2          )                          n          +          1                                      (          x          +          2          )                n              =          (      x      +      2      )                          2        n                    2                  n          +          1                      =          1      2      So, factoring the             1      2       and             (      x      +      2      )       outside, maintaining its absolute value, we get            1      2              |      x      +      2      |              lim              n        →        ∞                            ln                  (          n          )                            ln                  (          n          +          1          )                    . (We drop the absolute values on the logarithms as we&#039;re heading toward infinity -- everything is positive)            lim              n        →        ∞                            ln                  (          n          )                            ln                  (          n          +          1          )                      =          ∞      ∞       -- indeterminate, we should quickly apply l&#039;Hospital&#039;s Rule, rewriting with a hypothetical differentiable variable y, as n is not differentiable:                    lim              n        →        ∞                            ln        n                    ln                  (          n          +          1          )                      =          lim              y        →        ∞                            ln        y                    ln                  (          y          +          1          )                      =          lim              y        →        ∞                                      1          y                                      1                      y            +            1                                =          lim              y        →        ∞                            y        +        1            y        =    1  Therefore, we see             lim              n        →        ∞                            ln        n                    ln                  (          n          +          1          )                      =    1    ,   and we know we have convergence when       L    &amp;lt;    1   or            1      2              |      x      +      2      |        &amp;lt;    1              |      x      +      2      |        &amp;lt;    2    →    R    =    2  Then, determine the interval:      -    2    &amp;lt;    x    +    2    &amp;lt;    2        -    4    &amp;lt;    x    &amp;lt;    0     Test these endpoints:      x    =    0    :                      ∑                  n          =          2                ∞                                                  2            n                                                                          2              n                                                ln          n                      =                  ∑                  n          =          2                ∞                    1              ln        n            We&#039;ll use the Direct Comparison Test:             1      n        ≤          1              ln                  (          n          )                     on             [      2      ,      ∞      )      , we know this because the logarithm grows slower, and the smaller denominator ensures a larger sequence, thus, since the smaller series                     ∑                  n          =          2                ∞                    1      n       diverges by the p-series test where       p    =    1  , so does the larger series.As a result, this endpoint is excluded from the convergence interval.      x    =    -    4    :                      ∑                  n          =          2                ∞                                      (          -          2          )                n                              2          n                          ln          n                      =                  ∑                  n          =          2                ∞                                                  (            -            1            )                    n                                                    2              n                                                                                      2              n                                                ln          n                      =                  ∑                  n          =          0                ∞                                      (          -          1          )                n                    ln        n            Using the Alternating Series Test, we see              b      n        =          1              ln        n             is decreasing on             [      2      ,      ∞      )       due to the growing denominator and              lim              n        →        ∞                    1              ln        n              =    0  Thus, for this endpoint, we have convergence by the Alternating Series Test. Interval of Convergence:              [      -      4      ,      0      )

parlitz3s2 · Accepted Answer

lim              n        →        +        ∞                    |                        a                      n            +            1                                    a          n                    |        =          lim              n        →        +        ∞                    |                                                                  (                x                +                2                )                                            n                +                1                                                                    2                                  n                  +                  1                                                            ln                                  (                  n                  +                  1                  )                                                                                                                        (                x                +                2                )                                            n                                                                    2                                  n                                                            ln                                  (                  n                  )                                                                        |            =          lim              n        →        +        ∞                    |                        x          +          2                2                              ln          n                                      ln                          (              n              +              1              )                                          |            =          |                        x          +          2                2            |       as             lim              n        →        +        ∞                    |                        ln          n                                      ln                          (              n              +              1              )                                          |        =    1  The series converges for            |                        x          +          2                2            |        &amp;lt;    1        -    2    &amp;lt;    x    +    2    &amp;lt;    2        -    4    &amp;lt;    x    &amp;lt;    0  For       x    =    0                      ∑                  n          =          2                ∞                                      (          x          +          2          )                          n                                      2                      n                                    ln                      (            n            )                                =                  ∑                  n          =          2                ∞                            2        n                              2          2                          ln          n                      =                  ∑                  n          =          2                ∞                    1              ln        n            The series diverges for       x    =    0  For       x    =    -    4                      ∑                  n          =          2                ∞                                      (          -          4          +          2          )                          n                                      2                      n                                    ln                      (            n            )                                =                  ∑                  n          =          2                ∞                                      (          -          2          )                n                              2          2                          ln          n                      =                  ∑                  n          =          2                ∞                                      (          -          1          )                n                              ln          n                    This series converges conditionally for       x    =    -    4  The interval of convergence is       -    4    ≤    x    &amp;lt;    0

Find the radius of convergence and interval of convergence. sum_(n=2)^infty((x+2)^n)/(2^n ln n)?

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