# Evaluate $\displaystyle\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{a^n} + {a^{2n}}}}{{1 + a}}} \right)^{1/n}}$ where $0<a<1$

Since $0<a<1$, we have $a^{n}<a^{n}+a^{2n}<2a^{n}$, hence $a{\left( {\frac{1}{{1 + a}}} \right)^{1/n}} = {\left( {\frac{{{a^n}}}{{1 + a}}} \right)^{1/n}} < {\left( {\frac{{{a^n} + {a^{2n}}}}{{1 + a}}} \right)^{1/n}} < {\left( {\frac{{2{a^n}}}{{1 + a}}} \right)^{1/n}} = a{\left( {\frac{2}{{1 + a}}} \right)^{1/n}}$. But $\displaystyle\mathop {\lim }\limits_{n \to \infty } a{\left( {\frac{1}{{1 + a}}} \right)^{1/n}} = \displaystyle\mathop {\lim }\limits_{n \to \infty } a{\left( {\frac{2}{{1 + a}}} \right)^{1/n}} = a$, by Squeeze Theorem, $\displaystyle\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{a^n} + {a^{2n}}}}{{1 + a}}} \right)^{1/n}}=a.$

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Yeah this one looks good. You could even try the quotient criterium to check it. You can interpret you limit as the root test for the convergence of a series.

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You can evaluate the limit directly.

Since $\left( {\frac{{{a^n} + {a^{2n}}}}{{1 + a}}} \right)^{\frac{1}{n}} = a \left( {\frac{{1 + {a^{n}}}}{{1 + a}}} \right)^{\frac{1}{n}} = a \frac{(1 + {a^{n}})^{\frac{1}{n}}}{(1 + {a})^{\frac{1}{n}}} = a \frac{e^{\frac{1}{n} \ln(1+a^n)}}{e^{\frac{1}{n} \ln(1+a)}}$,

and $\lim_{n \to \infty} \ln(1+a^n) = 0$, and $\lim_{n \to \infty} \frac{1}{n} = 0$, we have $\lim_{n \to \infty} \left( {\frac{{{a^n} + {a^{2n}}}}{{1 + a}}} \right)^{\frac{1}{n}} = a$.

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