<s>[INST] How to evaluate $\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}}$? I need to resolve the convergence of $$\sum_{n = 1}^{\infty} \frac{n^3}{2^n + 5^n}$$ where $n \in \mathbb{N}$.
Mainly, I need advice concerning Cauchy's convergence criterion because I just can't evaluate
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}}.$$
My initial idea was to convert it to $\exp$ but that didn't help a bit:
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}} = \mathrm{e}^{\lim_{n \to \infty}\frac{1}{n}\ln\frac{n^3}{2^n + 5^n}}$$ but now I have to evaluate the bit 
$$\lim_{n \to \infty} \ln \frac{n^3}{2^n + 5^n}.$$
It clearly goes to $0$ since $2^n + 5^n$ grows much faster than $n^3$, but what now? $\ln 0 = -\infty$ and plugging it back will result in $0 \cdot \left(-\infty\right)$ which is undefied.
I didn't give up there. Let's take a different approach.
$$\lim_{n \to \infty} \sqrt[n]{\frac{n^3}{2^n + 5^n}} = \lim_{n \to \infty} \left(\frac{n^3}{2^n + 5^n}\right)^{\frac{1}{n}}$$
this will result in $\left(\frac{\infty}{\infty}\right)^0$, so maybe we can expand it:
$$\lim_{n \to \infty} \left(\frac{n^3}{2^n + 5^n}\right)^{\frac{1}{n}} = \lim_{n \to \infty}\frac{n^{\frac{3}{n}}}{\left(2^n + 5^n\right)^\frac{1}{n}} = \lim_{n \to \infty} \frac{e^{\frac{3}{n}\ln{n}}}{e^{\frac{1}{n} \ln\left(2^n + 5^n\right)}}$$ and it's still in an indetermined form. If I were to continue in this fashion I would, of course, get the same expression as before.
 [/INST] Just like started, first we switch to the exponent of $e$:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})$$
Using l'Hospitals rule we need to evaluate the derivative of the numerator:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{-1}\frac{3n^2(2^n+5^n)-n^3(2^n\ln2+5^n\ln5)}{(2^n+5^n)^2}=$$
$$\lim_{n\rightarrow\infty}\frac{3}{n}-\frac{2^n\ln2+5^n\ln5}{2^n+5^n}=\lim_{n\rightarrow\infty}\frac{3}{n}-\lim_{n\rightarrow\infty}\frac{(\frac{2}{5})^n\ln2+\ln5}{(\frac{2}{5})^n+1}=$$
$$0-\frac{0*\ln2+\ln5}{0+1}=-\ln5$$
Hence the result is:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})=\exp(-\ln5)=\frac{1}{5}$$
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