009C Sample Final 3, Problem 2

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Consider the series

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}}.}

(a) Test if the series converges absolutely. Give reasons for your answer.

(b) Test if the series converges conditionally. Give reasons for your answer.

Foundations:  
1. A series    is absolutely convergent if
        the series    converges.
2. A series    is conditionally convergent if
        the series    diverges and the series    converges.


Solution:

(a)

Step 1:  
First, we take the absolute value of the terms in the original series.
Let  
Therefore,
       
Step 2:  
This series is a  -series with  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p=1/2.}  
Therefore, it diverges.
Hence, the series
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}}
is not absolutely convergent.

(b)

Step 1:  
For
       
we notice that this series is alternating.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b_{n}={\frac {1}{\sqrt {n}}}.}
First, we have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {n}}}\geq 0}
for all  
The sequence    is decreasing since
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}}
for all  
Also,
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{\sqrt {n}}}=0.}
Therefore, the series     converges
by the Alternating Series Test.
Step 2:  
Since the series     is not absolutely convergent but convergent,
this series is conditionally convergent.


Final Answer:  
   (a)    not absolutely convergent (by the  -series test)
   (b)    conditionally convergent (by the Alternating Series Test)

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