prove that the series convereges and find its value $\frac{5^{k+1}+(-3)^k}{7^{k+2}}$

Question

$$\sum_{k=0}^{\infty}\frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$

My first inclination is to use the divergence test.

$$\lim_{k\to\infty} \frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$

$$ \frac{5^{\infty}+(-3)^{k}}{7^{\infty}} \to \frac{5^{\infty}+(-3)^{\infty}}{7^{\infty}}$$

$$\frac{5^{\infty}+(-3)^{k}}{7^{\infty}} \to \frac{\infty+\infty}{\infty}$$

Since the sequence $a_k$ does not converge the series diverges by the diveregence test

Answer 1

hint

We have

$$|5^{k+1}+(-3)^k | \le 5^{k+1}+3.5^{k+1}\le 5^{k+2}$$

thus $$|u_n| \le (\frac 57)^{k+2}$$

Its sum is

$$\frac{5}{49}\sum_{k=0}^\infty (\frac 57)^k+\frac{1}{49}\sum_{k=0}^\infty(\frac{-3}{7})^k=$$ $$\frac{5}{49}\frac{1}{1-\frac 57}+\frac{1}{49}\frac{1}{1+\frac{3}{7}}=\frac{13}{35}$$