Find the limit of $\lim_{n\to \infty} \sum_{x=n}^\infty \frac{1}{x^3}$ Find the limit of $\lim_{n\to \infty} \sum_{x=n}^\infty \frac{1}{x^3}$. Im having issues understanding how to prove this since I have never done something like this. I want to say it goes to zero since I know the series itself converges meaning it also tends to zero. Since x scales with n, once n is sufficiently large the sum would also essentially be 0 is my thinking. I am just unsure of how to prove this rigorously.
 A: Your idea is correct: From the fact that a convergent sequence is also a Cauchy sequence we can conclude that the partial sums of the given series also has to be cauchy. Let
$$S_n = \sum_{x=1}^n \frac{1}{x^3}$$
then since we know that $\sum_{x=0}^{\infty} \frac{1}{x^3}$ converges (meaning the sequence $(S_n)_n$ converges for $n \to \infty$) we know that for every $\epsilon >0$ there exist sufficiently large $n,m \in \mathbb N$ where (WLOG) $n \leq m$:
$$|S_n-S_m| < \epsilon \Leftrightarrow |\sum_{x=n+1}^m \frac{1}{x^3}| < \epsilon$$
In this expression we can use that all the components are positive, so $|\sum_{x=n+1}^m \frac{1}{x^3}| = \sum_{x=n+1}^m \frac{1}{x^3} < \epsilon$ and by letting $m$ tend to $\infty$ (since this inequality holds true for all $n,m$ sufficiently large, it doesn't matter if $n$ and $m$ are close to one another) we obtain
$$\epsilon \ge \lim_{m \to \infty} \sum_{x=n+1}^m \frac{1}{x^3} = \sum_{x=n+1}^{\infty} \frac{1}{x^3}$$
for sufficiently large $n$. Since $\epsilon>0$ was chosen arbitrarily we get that
$$\lim_{n \to \infty} \sum_{x=n+1}^{\infty} \frac{1}{x^3} =0$$
which means the statement is true.
A: If you already know that a series $\displaystyle\sum_{n=m}^\infty a_n$ converges, then you can write
$$\sum_{n=m}^\infty a_n = \sum_{n=1}^\infty a_n - \sum_{n=1}^{m-1} a_n$$
Taking limits on both sides we have $$\lim_{m\to\infty}\sum_{n=m}^\infty a_n = \sum_{n=1}^\infty a_n - \lim_{m\to\infty}\sum_{n=1}^{m-1} a_n =  \sum_{n=1}^\infty a_n - \sum_{n=1}^\infty a_n = 0$$
