I would like to ask why the following modification of the example is not allowed $$\displaystyle\lim_{n\rightarrow \infty} \frac{1+2+ \cdots + 2^n}{1+5+ \cdots + 5^n} = \lim_{n\to \infty}\left(\frac{2^n}{5^n}\right)\cdot \frac{\left(\frac{1}{2^n}+\frac{2}{2^n}+.....+1\right)}{\left(\frac{1}{5^n}+\frac{5}{5^n}+.....+1\right)}$$ and similarly
$$\displaystyle\lim _{n\rightarrow \infty} \left(\frac{3n+1}{4n+5}\right) ^n = \lim_{n \to \infty} \left(\frac{3}{4}\right)^n\cdot\left(\frac{1 + \frac{1}{3n}}{1 + \frac{5}{4n}}\right)^n$$
For these two examples, I am not concerned with the calculation, but rather with describing why this procedure is wrong.
To actually evaluate the limits I'd go as follows:
$$\frac{1+2+\ldots+2^n}{1+5+\ldots+5^n}=\frac{\frac{1-2^{n+1}}{1-2}}{\frac{1-5^{n+1}}{1-5}}=4\frac{1-2^{n+1}}{1-5^{n+1}}=4\left(\frac25\right)^{n+1}\;\frac{\frac1{2^{n+1}}-1}{\frac1{5^{n+1}}-1}\xrightarrow[n\to\infty]{}4\cdot0\cdot1=0$$
As for the second one:
$$\left(\frac{3n+1}{4n+5}\right)^n=\frac1{\left(\frac43\right)^n}\left(\frac{3n+1}{3n+\frac{15}4}\right)^n=\frac1{\left(\frac43\right)^n}\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^n=$$
$$=\frac1{\left(\frac43\right)^n}\left[\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^{3n+\frac{15}4}\right]^{-2n-\frac{15}4}\xrightarrow[n\to\infty]{}0\cdot0=0$$