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.
 A: Observe that $2^n+2^n<2^n+5^n<5^n+5^n$, $n^{3/n}$ goes to $1$ for large $n$ and $\frac{n+1}{n}$ goes to $1$ for large $n$. Using the roots test you can squeeze your answer to $2$ favorable answers and draw a conclusion
A: 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:
$$\lim_{n\rightarrow\infty}(\frac{n^3}{2^n+5^n})^{1/n} = \exp{(\lim_{n\rightarrow\infty}\frac{\ln(\frac{n^3}{2^n+5^n})}{n}})=\exp(-\ln5)=\frac{1}{5}$$
A: Lemma: Let $p(x)=a_kx^k+\dots+a_1x+a_0$ a polynomial of degree $k$. Then $\lim_{n\to\infty}\sqrt[n]{|p(n)|}=1.$
Proof: $|p(n)|=|a_kn^k+\dots+a_1n+a_0|=n^k\left|a_k+\frac{a_{k-1}}{n}+\dots+\frac{a_0}{n^k}\right|$.
Hence, 
$$\sqrt[n]{|p(n)|}=(n^{1/n})^k\left|a_k+\frac{a_{k-1}}{n}+\dots+\frac{a_0}{n^k}\right|^{1/n}.$$
Taking the limit and using that $\lim_{n\to\infty} n^{1/n}=1$, we get
$$\lim_{n\to\infty}\sqrt[n]{|p(n)|}=(1)^k|a_k|^{0}=1.$$
Theorem: Let $a>1$ and $p(x)=a_kx^k+\dots+a_1x+a_0$  a polynomial of degree $k$.  Then, $\sum_{n=1}^\infty\frac{p(n)}{a^n}$ is absolutely convergent, in particular convergent. 
Proof: Bythe previous lemma, we have $\lim_{n\to\infty}\sqrt[n]{\left|\frac{p(n)}{a^n}\right|}=\lim_{n\to\infty}\frac{\sqrt[n]{|p(n)|}}{a}=\frac{1}{a}<1$. Hence, by Ratio test, we have that $\sum_{n=1}^\infty\frac{p(n)}{a^n}$ is absolutely convergent. 
Going back to the problem
In the original problem you actually had something of the form $\sum \frac{p(n)}{a^n+b^n}$, with $a<b$ and $b>1$. this is also absolutely convergent. You can do it by comparison to the series $\sum \frac{p(n)}{b^n}$.
Alternatively, you can do it by ratio test again: Note
$\frac{|p(n)|}{b^n+b^n}<\frac{|p(n)|}{a^n+b^n}<\frac{|p(n)|}{b^n}$
So, 
$\sqrt[n]{\frac{|p(n)|}{b^n+b^n}}<\sqrt[n]{\frac{|p(n)|}{a^n+b^n}}<\sqrt[n]{\frac{|p(n)|}{b^n}}$. And taking the limit, by the squeeze theorem, we get $\lim_{n\to\infty}\sqrt[n]{\frac{|p(n)|}{a^n+b^n}}=\frac{1}{b}<1$. 
A: When using the root test,
when there are powers
(such as the $2^n$ and $5^n$ here)
polynomials don't matter
(except at the edge)
and only the largest 
base matters,
since the others
get swallowed up.
In this case,
all that matters is
the $5^n$,
so that the root test gives
$\dfrac1{5}$;
since this is less than
$1$,
the sum converges.
