Does the series diverge or converge and find the sum if possible.

Question

Does the series diverge or converge?

$$\sum_{n=1}^\infty \frac{3}{5^n - e^n}$$

Answer 1

By the Ratio Test we have $$L=\lim_{n\rightarrow \infty} \frac{\frac{3}{5^{n+1}-e^{n+1}}}{\frac{3}{5^n-e^n}}=\lim_{n\rightarrow \infty}\frac{5^n-e^n}{5^{n+1}-e^{n+1}}=\lim_{n\rightarrow \infty}\frac{1-\frac{e^n}{5}}{5-e\frac{e^n}{5^n}}$$

Since $\lim_{n\rightarrow \infty}e\frac{e^n}{5^n}=\lim_{n\rightarrow \infty}e(\frac{e}{5})^n=0$ it follows that

$L=\frac15\lt1$, thus the series is convergant.

Answer 2

hint: $5^n - e^n = e^n\left(\left(\dfrac{5}{e}\right)^n - 1\right)> e^n, n \geq 2$

Answer 3

Use the ratio test:

$$\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_{n}} = \frac{5^{n} - e^{n}}{5^{n+1} - e^{n+1}} = \frac{\frac{5^{n}}{5^{n+1}} - \frac{e^{n}}{5^{n+1}}}{1 - \frac{e^{n+1}}{5^{n+1}}} \rightarrow \frac{\frac{1}{5} - 0}{1 - 0} = \frac{1}{5}$$

and go from there. Since your previous answer is a hint, this is too.