I have the infinte series $\sum_{n=1}^{\infty}\frac{1}{n^3}$ which I believe converges.
As the ratio test proved inconclusive, I am trying to use the comparison test in order to prove it's convergence, but I am unsure what series to compare it to. Initially, I believed I could compare it to $\frac{1}{n^2}$ but I am also unsure how to prove that this series converges.
Is there a general method for choosing the series with which you compare your series to?
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You can use that $$\frac{1}{n^3}\le \frac{1}{n^2}$$ and $$\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$$
On
$$\sum_{n=1}^\infty \frac1{n^s} \lt 1+ \int_1^\infty \frac1{x^s} dx=\frac{s}{s-1}$$ For all $s\gt 1$.
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You can also use integral test i.e testing whether $\int_{1}^{\infty}{\frac{dx}{x^3}}$ converges which seem to be easy to solve
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The answer is Apery's constant, or the limit of the Zeta function at $s=3$
https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
There is no known way to evaluate this, but it does converge, as it is smaller than the series for $s=2$, whose proof is given in this beautiful video by 3Blue1Brown
Yes, you can use the fact that, for each natural number $n$, $\dfrac1{n^3}\leqslant\dfrac1{n^2}$. And then you can use the fact that $\dfrac1{n^2}\leqslant\dfrac1{n(n-1)}$ if $n\neq1$. Now, you have$$\sum_{n=2}^\infty\frac1{n(n-1)}=\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1n\right),$$and this series converges (it is a telescoping series).
But it is more natural to use the integral test to study the convergence of $\displaystyle\sum_{n=1}^\infty\frac1{n^3}$.