Calculus BAlternating Series Problems 1

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