009C Sample Final 3, Problem 4 Detailed Solution

Determine if the following series converges or diverges. Please give your reason(s).

(a)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }{\frac {n!}{(2n)!}}}

(b)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}}}

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Background Information:
1. Ratio Test
Let  $\sum a_{n}$  be a series and  $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.$
Then,
If  $L<1,$  the series is absolutely convergent.
If  $L>1,$  the series is divergent.
If  $L=1,$  the test is inconclusive.
2. If a series absolutely converges, then it also converges.
3. Alternating Series Test
Let  $\{a_{n}\}$  be a positive, decreasing sequence where  $\lim _{n\rightarrow \infty }a_{n}=0.$
Then,  $\sum _{n=1}^{\infty }(-1)^{n}a_{n}$  and  $\sum _{n=1}^{\infty }(-1)^{n+1}a_{n}$
converge.

Solution:

(a)

Step 1:
We begin by using the Ratio Test.
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)!}{(2(n+1))!}}{\frac {(2n)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n!}{(2n+2)(2n+1)(2n)!}}{\frac {(2n)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n+1}{(2n+2)(2n+1)}}}\\&&\\&=&\displaystyle {0.}\end{array}}}

Step 2:
Since
$\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}=0<1,$
the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

Step 1:
For
$\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}},$
we notice that this series is alternating.
Let  $b_{n}={\frac {1}{n+1}}.$
First, we have
${\frac {1}{n+1}}\geq 0$
for all  $n\geq 1.$
The sequence  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 \{b_n\}}   is decreasing since
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}{n+2}<\frac{1}{n+1}}
for all  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 n\ge 1.}

Step 2:
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+1}=0.}
Therefore,
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 (-1)^n\frac{1}{n+1}}
converges by the Alternating Series Test.

Final Answer:
(a)    converges (by the Ratio Test)
(b)    converges (by the Alternating Series Test)

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