series 1 lecture participation (2 points)\nlet\nfor the following answer blanks, decide whether the given…
series 1 lecture participation (2 points)\nlet\nfor the following answer blanks, decide whether the given sequen to ∞, -infinity if it diverges to -∞ or dne otherwise.\n(a) the series $sum_{n = 1}^{infty}\frac{3n}{11 - 2n}$.\n(b) the sequence ${\frac{3n}{11 - 2n}}$.\nnote: you can earn partial credit on this problem.\npreview my answers\nsubmit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.
Answer
Explanation:
Step1: Analyze the series $\sum_{n = 1}^{\infty}\frac{3n}{11-2n}$
Use the n - th term test for divergence. If $\lim_{n\rightarrow\infty}a_{n}\neq0$, then $\sum_{n = 1}^{\infty}a_{n}$ diverges. Calculate $\lim_{n\rightarrow\infty}\frac{3n}{11 - 2n}$. Divide both numerator and denominator by $n$: $\lim_{n\rightarrow\infty}\frac{3n/n}{(11 - 2n)/n}=\lim_{n\rightarrow\infty}\frac{3}{\frac{11}{n}-2}$. As $n\rightarrow\infty$, $\frac{11}{n}\rightarrow0$. So $\lim_{n\rightarrow\infty}\frac{3}{\frac{11}{n}-2}=\frac{3}{0 - 2}=-\frac{3}{2}\neq0$. Since the limit of the general term is not zero, the series diverges, and it does not diverge to $\pm\infty$.
Step2: Analyze the sequence $\left{\frac{3n}{11-2n}\right}$
Again, divide both numerator and denominator by $n$: $\lim_{n\rightarrow\infty}\frac{3n/n}{(11 - 2n)/n}=\lim_{n\rightarrow\infty}\frac{3}{\frac{11}{n}-2}$. As $n\rightarrow\infty$, $\frac{11}{n}\rightarrow0$. So $\lim_{n\rightarrow\infty}\frac{3}{\frac{11}{n}-2}=\frac{3}{0 - 2}=-\frac{3}{2}$. The sequence converges to $-\frac{3}{2}$, but we are asked for divergence - related answers. Since it does not diverge to $\pm\infty$, the answer is DNE.
Answer:
(a) DNE (b) DNE