009C Sample Midterm 2, Problem 3

Determine convergence or divergence:

(a) $\sum _{n=1}^{\infty }(-1)^{n}{\sqrt {\frac {1}{n}}}$

(b) $\sum _{n=1}^{\infty }(-2)^{n}{\frac {n!}{n^{n}}}$

Foundations:
1. 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.
2. 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.
3. If a series absolutely converges, then it also converges.

Solution:

(a)

Step 1:
First, we have
$\sum _{n=1}^{\infty }(-1)^{n}{\sqrt {\frac {1}{n}}}=\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{\sqrt {n}}}.$

Step 2:
We notice that the series is alternating.
Let $b_{n}={\frac {1}{\sqrt {n}}}.$
The sequence $\{b_{n}\}$ is decreasing since
${\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}$
for all $n\geq 1.$
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}{\sqrt{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}{\sqrt{n}}} converges by the Alternating Series Test.

(b)

Step 1:
We begin by 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{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(-2)^{n+1} (n+1)!}{(n+1)^{n+1}} \frac{n^n}{(-2)^n n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\ &&\\ & = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.} \end{array}}

Step 2:
Now, we need to calculate 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} \bigg(\frac{n}{n+1}\bigg)^n.}
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 y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
Then, taking the natural log of both sides, we get
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{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}} \end{array}}
since we can interchange limits and continuous functions.
Now, this limit has the form 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{0}{0}.}
Hence, we can use L'Hopital's Rule to calculate this limit.

Step 3:
Now, 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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\bigg(\frac{x}{x+1}\bigg)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\ &&\\ & = & \displaystyle{-1.} \end{array}}

Step 4:
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 \ln y=-1,} we know
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 y=e^{-1}.}
Now, 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 \lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.}
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{2}{e}<1,} the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

Final Answer:
(a)     converges
(b)     converges

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