Checking convergence of seriesMy task is to check whether the series ∑ n = 1 ∞ ( − 2 ) n + n 2 n 2 n converges or not. So I tried expanding the fraction like that(should equal to original form): ∑ n = 1 ∞ ( − 1 ) n 2 n n 2 n + n 2 n 2 n = ∑ n = 1 ∞ ( − 1 ) n 1 n + n 2 n Then I thought about that the first addend does converge because of Leibniz's rule and the second addend converges because 2 n grows much faster than nBut I'm not sure(and rather thinking this form of argumentation is wrong) if I can take addends of a series and prove convergence for each addend instead of proving convergence for the whole series.I would appreciate if someone can point me to the right direction.

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Checking convergence of seriesMy task is to check whether the series      ∑          n      =      1        ∞                      (        −        2                  )          n                +                  n          2                            n                  2          n                    converges or not. So I tried expanding the fraction like that(should equal to original form):      ∑          n      =      1        ∞        (    −    1          )      n                      2        n                    n                  2          n                      +                  n        2                    n                  2          n                      =      ∑          n      =      1        ∞        (    −    1          )      n              1      n        +          n              2        n            Then I thought about that the first addend does converge because of Leibniz&#039;s rule and the second addend converges because       2    n   grows much faster than   nBut I&#039;m not sure(and rather thinking this form of argumentation is wrong) if I can take addends of a series and prove convergence for each addend instead of proving convergence for the whole series.I would appreciate if someone can point me to the right direction.

Stevinivm · Accepted Answer

The 1st one is   −  ln  ⁡  2 (Taylor series).The 2nd one is  f  (  x  )  =      ∑          n      =      1              ∞        n  ⋅      x    n    ∫            f      (      x      )        x    d  x  =      ∑          n      =      1              ∞            x    n    =      x          1      −      x        ∴  f  (  x  )  =  x  ⋅            (              x                  1          −          x                    )        ′    =      x          (      1      −      x              )        2                  ∑          n      =      1              ∞        n  ⋅            (              1        2            )        n    =            1      2              (      1      −              1        2                    )        2              =  2So the total sum is   2  −  ln  ⁡  2

boloman0z · Accepted Answer

To control if the series:      ∑          n      =      1        ∞        (    −    1          )      n              1      n        +          n              2        n            you can use the Leibniz&#039;s rule. First you have to demonstrate that  −      1    n    +      n          2      n        &amp;lt;  0for every   n. Indeed you can note that rewriting the expression  −                    2        n            +              n        2                    n      ⋅              2        n              &amp;lt;  0. Therefore now you know that for   n  →  ∞      1    n    +      n          2      n        →  0therefore you have to demonstrate      1    n    +      n          2      n      is decreasing proving that the derivate of   f  (  x  )  =      1    x    +      x          2      x        &amp;lt;  0 after a value of   x.

Checking convergence of series My task is to check whether the series <munderover> &#x22

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