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`sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` Determine whether the series converges absolutely or conditionally, or diverges.

To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` , we may apply Alternating Series Test.


In Alternating Series Test, the series `sum (-1)^(n+1) a_n ` is convergent if:


1) `a_ngt=0`


2) ` a_n` is monotone and decreasing sequence.


3) `lim_(n-gtoo) a_n =0`


For the series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` , we have:


`a_n = 1/(nsqrt(n))`


Apply the radical property: `sqrt(x) =x^(1/2)` and Law of Exponents: `x^n*x^m =x^(n+m).`


`a_n = 1/(nsqrt(n))`


      `=1/(n*n^(1/2))`


      `=1/n^(1+1/2)`


      `=1/n^(3/2)`


The `a_n =1/n^(3/2) ` is a decreasing sequence.


Then, we set-up the limit as :


`lim_(n-gtoo)1/n^(3/2) = 1/oo =0`


By alternating series test criteria, the series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` converges.


The series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` has positive and negative elements. Thus, we must verify if the series converges absolutely or conditionally. Recall:


a) Absolute Convergence:  `sum a_n`  is absolutely convergent if `sum|a_n|`  is convergent.  


b) Conditional Convergence:  `sum a_n`  is conditionally convergent if `sum|a_n|`   is divergent and `sum a_n`  is convergent.  


We evaluate the `sum |a_n` | as :


`sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))| =sum_(n=1)^oo 1/(nsqrt(n))`


                         `=sum_(n=1)^oo 1/n^(3/2)`


Apply the p-series test `sum_(n=1)^oo 1/n^p` is convergent if `pgt1` and divergent if `0ltplt=1` .


The series `sum_(n=1)^oo 1/n^(3/2)` has `p=3/2 or 1.5` which satisfies `pgt1` . Thus, the `sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))|` is convergent.


Conclusion:


Based on Absolute convergence criteria, the series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` is absolutely convergent since  `sum |a_n| ` as `sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))|` is convergent.

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