Question

determine convergence or divergence of $sum_{n = 1}^{infty}\frac{14^{n}}{15^{n}-2n}$ using any method covered so far. the limit $l=lim_{n
ightarrowinfty}a_{n}=0$, where $a_{n}=\frac{14^{n}}{15^{n}-2n}$. by the $n$th term divergence test, the series converges. the limit $l=lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}} = 0$, 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 $l=lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}>0$, 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 $l=lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}>0$, 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.

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 \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}\).

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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 \(n
ightarrow\infty\), \(\lim_{n
ightarrow\infty}\ln y=-\infty\), so \(\lim_{n
ightarrow\infty}y = 0\). Then \(\lim_{n
ightarrow\infty}\frac{1}{1 - \frac{2n}{15^{n}}}=1>0\).

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 \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}} = 1>0\) 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 \(L=\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}>0\), 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.