determine convergence or divergence of $sum_{n = 1}^{infty}\frac{14^{n}}{15^{n}-2n}$ using any method…

determine convergence or divergence of $sum_{n = 1}^{infty}\frac{14^{n}}{15^{n}-2n}$ using any method covered so far. the limit $l=lim_{n\rightarrowinfty}a_{n}=0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$. by the $n$th term divergence test, the series converges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}} = 0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series diverges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}}>0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series diverges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}}>0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series converges. the infinite series is a geometric series with $|r|<1$. thus, the series converges.

determine convergence or divergence of $sum_{n = 1}^{infty}\frac{14^{n}}{15^{n}-2n}$ using any method covered so far. the limit $l=lim_{n\rightarrowinfty}a_{n}=0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$. by the $n$th term divergence test, the series converges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}} = 0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series diverges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}}>0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series diverges. the limit $l=lim_{n\rightarrowinfty}\frac{a_{n}}{b_{n}}>0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$ and $b_{n}=\frac{14^{n}}{15^{n}}$. by the limit comparison test, the series converges. the infinite series is a geometric series with $|r|<1$. thus, the series converges.

Answer

Explanation:

Step1: Apply Limit - Comparison Test

Let (a_{n}=\frac{14^{n}}{15^{n}-2n}) and (b_{n}=\frac{14^{n}}{15^{n}}). Then, find (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}). [ \begin{align*} \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}&=\lim_{n\rightarrow\infty}\frac{\frac{14^{n}}{15^{n}-2n}}{\frac{14^{n}}{15^{n}}}\ &=\lim_{n\rightarrow\infty}\frac{14^{n}\times15^{n}}{14^{n}(15^{n}-2n)}\ &=\lim_{n\rightarrow\infty}\frac{15^{n}}{15^{n}-2n}\ &=\lim_{n\rightarrow\infty}\frac{1}{1 - \frac{2n}{15^{n}}} \end{align*} ]

Step2: Evaluate the limit of (\frac{2n}{15^{n}})

We use L'Hopital's rule. Let (y = \frac{2n}{15^{n}}), then (\ln y=\ln(2n)-n\ln15). As (n\rightarrow\infty), (\lim_{n\rightarrow\infty}\ln y=-\infty), so (\lim_{n\rightarrow\infty}y = 0). Then (\lim_{n\rightarrow\infty}\frac{1}{1 - \frac{2n}{15^{n}}}=1>0).

Step3: Analyze the convergence of (\sum_{n = 1}^{\infty}b_{n})

The series (\sum_{n = 1}^{\infty}b_{n}=\sum_{n = 1}^{\infty}(\frac{14}{15})^{n}) is a geometric series with common - ratio (r=\frac{14}{15}) and (|r|=\frac{14}{15}<1), so (\sum_{n = 1}^{\infty}b_{n}) converges.

Step4: Use the Limit - Comparison Test conclusion

Since (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}} = 1>0) and (\sum_{n = 1}^{\infty}b_{n}) converges, by the Limit - Comparison Test, (\sum_{n = 1}^{\infty}a_{n}=\sum_{n = 1}^{\infty}\frac{14^{n}}{15^{n}-2n}) converges.

Answer:

The limit (L=\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}>0), where (a_{n}=\frac{14^{n}}{15^{n}-2n}) and (b_{n}=\frac{14^{n}}{15^{n}}). By the Limit Comparison Test, the series converges.