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Alternating Series:
\( \sum_{n=1}^{\infty}(-1)^{n} b n \) \( n=1 \quad(-1)^1=-1 \) \( -b_{1}+b_{2}-b_{3}+b_4\cdots \)
\( \sum_{n=1}^{\infty}(-1)^{n-1} b n \) \( n=1 \quad(-1)^{0}=1 \) \( b_{1}-b_{2}+b_{3}-b_{4} \dots \)
Test for Alternating series: is convergent if:
① \( \lim _{n \rightarrow \infty} b_{n}=0 \)
② Decreasing \( \Rightarrow b_{n+1} \leq b_{n} \)
Test the series \( \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n}\right) \) for convergence.
A.S.T:
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{1}{x} \)
\( b_{n+1}=\frac{1}{n+1}<b n=\frac{1}{n} \)
\( \sum(-1)^{n-1}\left(\frac{1}{n}\right) \) is Convergent by A.S.T
Test the series \( \sum(-1)^{n} \frac{\ln (n)}{n} \) for convergence.
The sries is alternating \( \Rightarrow \) Apply A.S.T:
\( b n=\frac{\ln (n)}{n} \)
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\frac{\infty}{\infty} \) use L'Hopital Rule
\( \lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\lim _{n \rightarrow \infty} \frac{1/n}{1}=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{\ln (x)}{x} \Rightarrow f^{\prime}(x)=\frac{\frac{1}{x} \cdot x-(1)(\ln (x))}{x^{2}}=\frac{1-\ln (x)}{x^{2}} \)
\( x^{2} \Rightarrow \) is always positive, \( \ln (e)=1 \) \( , \ln (n) >1, n=3,4 \cdots \)
\( ,\ln(n)< 1, n=1,2 \)
\( 1-\ln (x)<0 \) \( \Rightarrow \) for \( x \geq 3 \)
\( \{b n\} \) is decreasing
\( \sum(-1)^{n} \frac{\ln (n)}{n} \) is Convergent by A.S.T
Test the series \( \sum \frac{\cos (n \pi)}{n^{2}+1} \) for convergence.
Alternating Series
\( \sum_{n=1}^{\infty}(-1)^{n} b n \)
\( b_{n}=\frac{1}{n^{2}+1} \)
\( \sum_{n=1}^{\infty} \cos (n \pi)=\cos (\pi)+\cos (2 \pi)+\cos (3 \pi) \ldots \)
\( =-1+1-1+1 \)
\( =\sum_{n=1}^{\infty}(-1)^{n} \)
\( \sum \frac{\cos (n \pi)}{n^{2}+1}=\sum \frac{(-1)^{n}}{n^{2}+1} \) Alternating series
\( b n=\frac{1}{n^{2}+1} \)
Apply A.S.T \( \Rightarrow \)
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n^{2}+1}=\frac{1}{\infty^{2}}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{1}{x^{2}+1} \Rightarrow f^{1}(x)=\frac{-2 x(1)}{\left(x^{2}+1\right)^{2}}<0 \)
\( \left\{b{n}\right\} \) is decreasing
\( \sum \frac{\cos (n \pi)}{n^{2}+1} \) is Convergent by A.S.T
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Alternating Series:
\( \sum_{n=1}^{\infty}(-1)^{n} b n \) \( n=1 \quad(-1)^1=-1 \) \( -b_{1}+b_{2}-b_{3}+b_4\cdots \)
\( \sum_{n=1}^{\infty}(-1)^{n-1} b n \) \( n=1 \quad(-1)^{0}=1 \) \( b_{1}-b_{2}+b_{3}-b_{4} \dots \)
Test for Alternating series: is convergent if:
① \( \lim _{n \rightarrow \infty} b_{n}=0 \)
② Decreasing \( \Rightarrow b_{n+1} \leq b_{n} \)
Test the series \( \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n}\right) \) for convergence.
A.S.T:
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{1}{x} \)
\( b_{n+1}=\frac{1}{n+1}<b n=\frac{1}{n} \)
\( \sum(-1)^{n-1}\left(\frac{1}{n}\right) \) is Convergent by A.S.T
Test the series \( \sum(-1)^{n} \frac{\ln (n)}{n} \) for convergence.
The sries is alternating \( \Rightarrow \) Apply A.S.T:
\( b n=\frac{\ln (n)}{n} \)
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\frac{\infty}{\infty} \) use L'Hopital Rule
\( \lim _{n \rightarrow \infty} \frac{\ln (n)}{n}=\lim _{n \rightarrow \infty} \frac{1/n}{1}=\lim _{n \rightarrow \infty} \frac{1}{n}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{\ln (x)}{x} \Rightarrow f^{\prime}(x)=\frac{\frac{1}{x} \cdot x-(1)(\ln (x))}{x^{2}}=\frac{1-\ln (x)}{x^{2}} \)
\( x^{2} \Rightarrow \) is always positive, \( \ln (e)=1 \) \( , \ln (n) >1, n=3,4 \cdots \)
\( ,\ln(n)< 1, n=1,2 \)
\( 1-\ln (x)<0 \) \( \Rightarrow \) for \( x \geq 3 \)
\( \{b n\} \) is decreasing
\( \sum(-1)^{n} \frac{\ln (n)}{n} \) is Convergent by A.S.T
Test the series \( \sum \frac{\cos (n \pi)}{n^{2}+1} \) for convergence.
Alternating Series
\( \sum_{n=1}^{\infty}(-1)^{n} b n \)
\( b_{n}=\frac{1}{n^{2}+1} \)
\( \sum_{n=1}^{\infty} \cos (n \pi)=\cos (\pi)+\cos (2 \pi)+\cos (3 \pi) \ldots \)
\( =-1+1-1+1 \)
\( =\sum_{n=1}^{\infty}(-1)^{n} \)
\( \sum \frac{\cos (n \pi)}{n^{2}+1}=\sum \frac{(-1)^{n}}{n^{2}+1} \) Alternating series
\( b n=\frac{1}{n^{2}+1} \)
Apply A.S.T \( \Rightarrow \)
① \( \lim _{n \rightarrow \infty} b n=\lim _{n \rightarrow \infty} \frac{1}{n^{2}+1}=\frac{1}{\infty^{2}}=\frac{1}{\infty}=0 \)
② \( f(x)=\frac{1}{x^{2}+1} \Rightarrow f^{1}(x)=\frac{-2 x(1)}{\left(x^{2}+1\right)^{2}}<0 \)
\( \left\{b{n}\right\} \) is decreasing
\( \sum \frac{\cos (n \pi)}{n^{2}+1} \) is Convergent by A.S.T
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