009C Sample Midterm 2, Problem 5

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If    converges, does it follow that the following series converges?

(a)  

(b)  


Foundations:  
Ratio Test
        Let    be a series and  
        Then,

        If    the series is absolutely convergent.

        If    the series is divergent.

        If    the test is inconclusive.


Solution:

(a)

Step 1:  
Assume that the power series    converges.
Let    be the radius of convergence of this power series.
We can use the Ratio Test to find   
Using the Ratio Test, we have

       

Since the radius of convergence of the series    is  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 R,}   we have
        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 R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.}
Step 2:  
Now, we use the Ratio Test to find the radius of convergence of 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=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.}
Using the Ratio Test, we have
        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 \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}2^nx^{n+1}}{c_n2^{n+1}x^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{2c_n}\bigg|}\\ &&\\ & = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).} \end{array}}
Hence, the radius of convergence of this power series is
        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 \frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=2R.}
Therefore, this power series converges.

(b)

Step 1:  
Assume that the power 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=0}^\infty c_nx^n}   converges.
Let  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 R}   be the radius of convergence of this power series.
We can use the Ratio Test to find  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 R.}  
Using the Ratio Test, we have

        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 \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).} \end{array}}

Since the radius of convergence of 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=0}^\infty c_nx^n}   is  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 R,}   we have
        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 R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.}
Step 2:  
Now, we use the Ratio Test to find the radius of convergence of 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=0}^\infty c_n(-x)^n .}
Using the Ratio Test, we have
        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 \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}(-x)}{c_n}\bigg|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).} \end{array}}
Hence, the radius of convergence of this power series is
        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 \frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=R.}
Therefore, this power series converges.


Final Answer:  
    (a)     converges
    (b)     converges

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