$ \sum_{n=0}^\infty ((\frac{n^3+2n+1}{n^3+n+1})^{1/3}-1)^{3/4} $
I created a graph and it seems like $ \frac{1}{n} $, but as i know this series should converge. Can't prove it with any usual rules, seems like i have to bound it with another converging function, but couldn't find anything better than $ \frac{1}{n}^{3/4} $, which is obviously don't converge.
$ ((1 +\frac{n}{n^3+n+1})^{1/3}-1)^{3/4} $ is like the most obvious step, but doesn't go that well afterwards. Any throughts?
On
We know that $$ \lim\limits_{x\to 0} \frac{e^{x/3}-1}{x} = 1/3 $$ So we can take $x=\ln(\frac{n^3+2n+1}{n^3+n+1}) $. And we know that $$ \lim\limits_{y\to 0} \frac{\ln(1+y)}{y} = 1 $$ So we can take $y=\frac{n+1}{n^3+n+1}$
Which we can compare further to $\frac{1}{n^2}$. Then by taking the power of $3/4$, we can compare the whole sum to $\frac{1}{n^{3/2}}$, so it's convergent
You can use Taylor expansions around $0$: as $n\to\infty$, $1/n\to 0$, and so $$\begin{align*} \frac{n^3+2n+1}{n^3+n+1} &= \frac{1+2/n^2+1/n^3}{1+1/n^2+1/n^3} = \frac{1+2/n^2+o(1/n^2)}{1+1/n^2+o(1/n^2)} \\&= \left(1+\frac{2}{n^2}+o\left(\frac{1}{n^2}\right)\right)\left(1-\frac{1}{n^2}+o\left(\frac{1}{n^2}\right)\right) \\&= 1+\frac{1}{n^2}+o\left(\frac{1}{n^2}\right) \end{align*}$$ (we used the Taylor expansion, to first order, of $\frac{1}{1-u}$ when $u\to 0$) from which $$\begin{align*} \left(\frac{n^3+2n+1}{n^3+n+1}\right)^{1/3} -1 &= \left(1+\frac{1}{n^2}+o\left(\frac{1}{n^2}\right)\right)^{1/3} -1 = 1+\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)-1 = \frac{1}{3n^2}+o\left(\frac{1}{n^2}\right) \end{align*}$$ (we used the Taylor expansion, to first order, of $(1+u)^\alpha$ when $u\to 0$) and thus $$\begin{align*} \left(\left(\frac{n^3+2n+1}{n^3+n+1}\right)^{1/3} -1\right)^{3/4} &= \frac{1}{3^{3/4}n^{3/2}}+o\left(\frac{1}{n^{3/2}}\right)\,. \end{align*}$$ You can now conclude by the limit comparison test, since you get $$ \boxed{\lim_{n\to\infty }\frac{\left(\left(\frac{n^3+2n+1}{n^3+n+1}\right)^{1/3} -1\right)^{3/4}}{\frac{1}{3^{3/4}n^{3/2}}} =1} $$ and the denominator is positive.