009C Sample Midterm 1, Problem 3

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Determine whether the following series converges absolutely,

conditionally or whether it diverges.

Be sure to justify your answers!


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:

Step 1:  
First, we take the absolute value of the terms in the original series.
Let
Therefore,
       
Step 2:  
This series is the harmonic series (or Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle p} -series with ).
Thus, it diverges. Hence, the series
       
is not absolutely convergent.
Step 3:  
Now, we need to look back at the original series to see
if it conditionally converges.
For
       
we notice that this series is alternating.
Let
The sequence is decreasing since
       
for all
Also,
        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 \lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.}
Therefore, 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=1}^\infty \frac{(-1)^n}{n}}   converges
by the Alternating Series Test.
Step 4:  
Since 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=1}^\infty \frac{(-1)^n}{n}}   is not absolutely convergent but convergent,
this series is conditionally convergent.


Final Answer:  
        Conditionally convergent

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