Question

7.) For the following strictly-non-negative - termed series state whether the series converges or diverges State which test you used to decide this convergence status. If you used either of the comparison tests, state to which series you compared your series. Use scratch paper to do your preliminary calculations. [6pts each] Series Convergence status Test applied to decide Σ0 3k 4k-1 0 4"-2" 4" ΣΗ-) +1 4"

Answer 1

Consider Series :  
"Consider Series: Sigma_k = 1^infinity (3k/4k - 1)^k Root Test: Consider a infinite term series Sigma a_k. let Lim_k rightarrow infinity

Root Test: Consider a infinite term series . Let , Then :

(i) if L<1 then series is convergent.

(ii) if L>1 then series is divergent

(iii) if L=1 then can not say anything means test fails.

Here ,  

Now,

  

  

  

Hence, by Root Test this series is Convergent.

Consider Series:

Ratio Test: Consider a infinite term series $\sum a_n$ . Let , Then :

(i) if L<1 then series is convergent.

(ii) if L>1 then series is divergent

(iii) if L=1 then can not say anything means test fails.

Here, $a_n=\frac{1}{4^n-2^n}$

So, $a_{n+1}=\frac{1}{4^{n+1}-2^{n+1}}$

Now,

$\lim_{n\rightarrow \infty }\left ( \frac{1}{4^{n+1}-2^{n+1}} \right .\frac{4^n-2^n}{1})$

$=\lim_{n\rightarrow \infty }\left ( \frac{4^n-2^n}{4^{n+1}-2^{n+1}} \right )$

$=\lim_{n\rightarrow \infty }\left ( \frac{4^n}{4^{n+1}-2^{n+1}} -\frac{2^n}{4^{n+1}-2^{n+1}}\right )$

$=\lim_{n\rightarrow \infty }\left ( \frac{1}{\frac{4^{n+1}}{4^n}-\frac{2^{n+1}}{4^{n}}} -\frac{1}{\frac{4^{n+1}}{2^n}-\frac{2^{n+1}}{2^{n}}}\right )$

$=\lim_{n\rightarrow \infty }\left ( \frac{1}{\frac{4^{n}.4}{4^n}-\frac{2^{n}.2}{4^{n}}} -\frac{1}{\frac{4^{n}.4}{2^n}-\frac{2^{n}.2}{2^{n}}}\right )$

   $=\lim_{n\rightarrow \infty }\left ( \frac{1}{4-\left ( \frac{1}{2} \right )^n.2} -\frac{1}{2^n.4-2}\right )$

   $=\frac{1}{4}-0$ { in first term 1/2 <1 so (1/2)^n will zero if n tends to infinity. In second term denominator will become infinity if n tends to infinity so whole term will be zero}

   $=L<1$

Hence, by Ratio Test Series is convergent.

Consider Series: $\sum_{n=1}^{\infty }\frac{n}{4^n}$

Here,   $a_n=\frac{n}{4^n}$

Now,

$\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left ( \frac{n+1}{4^{n+1}}.\frac{4^n}{n} \right )$

   $=\lim_{n\rightarrow \infty }\left ( \frac{n+1}{4^{n}.4}.\frac{4^n}{n} \right )$

$=\lim_{n\rightarrow \infty }\left ( \frac{n+1}{4}.\frac{1}{n} \right )$

   $=\lim_{n\rightarrow \infty }\left ( \frac{n+1}{4n} \right )$

$=\lim_{n\rightarrow \infty }\left ( \frac{1+\frac{1}{n}}{4} \right )$

$=L<1$

Hence by Ratio Test , series is convergent.

Consider Series:

Necessary Condition for convergence of a series $\sum a_n$ is that $\lim_{n\rightarrow \infty } a_n=0$ .

Here, $a_n=\left ( \frac{1}{4^n}+1 \right )$

Now,   $\lim_{n\rightarrow \infty }a_n=\lim_{n\rightarrow \infty }\left ( \frac{1}{4^n}+1 \right )={\color{Red} 1\neq 0}$

So, from above condition this series not convergent.

Final Result:

series1	Convergent	Root Test
series2	Convergent	Ratio Test
series3	Convergent	Ratio Test
series4	Not Convergent	Necessary Condition Test