Ratio test:
$$
\frac{\displaystyle\frac{5^{n+1}-2^{n+1}}{7^{n+1}-6^{n+1}}}{\displaystyle\frac{5^{n}-2^{n}}{7^{n}-6^{n}}}
=\frac{7^{n}-6^{n}}{{7^{n+1}-6^{n+1}}}\,\frac{5^{n+1}-2^{n+1}}{5^{n}-2^{n}}
=\frac{1-(6/7)^n}{7-(6/n)^{n+1}}\,\frac{5-(2/5)^{n+1}}{1-(2/5)^n}\to\frac57<1,
$$
so the series converges.
0
user180579
On
For each $n$, we have
$$0\le \frac{5^n-2^n}{7^n-6^n}\le \frac{5^n}{7^n-6^n}=\frac{(5/6)^n}{(7/6)^n - 1}.$$
Now use the ratio test on
$$\sum_{n=1}^\infty \frac{(5/6)^n}{(7/6)^n - 1}.$$
Ratio test: $$ \frac{\displaystyle\frac{5^{n+1}-2^{n+1}}{7^{n+1}-6^{n+1}}}{\displaystyle\frac{5^{n}-2^{n}}{7^{n}-6^{n}}} =\frac{7^{n}-6^{n}}{{7^{n+1}-6^{n+1}}}\,\frac{5^{n+1}-2^{n+1}}{5^{n}-2^{n}} =\frac{1-(6/7)^n}{7-(6/n)^{n+1}}\,\frac{5-(2/5)^{n+1}}{1-(2/5)^n}\to\frac57<1, $$ so the series converges.