Simple equivalent of the rest of the series $\sum\limits_n\frac1{n^3}$

Question

Consider the converging series \begin{equation} \sum_{n\geqslant1}{\frac{1}{n^3}} \end{equation} I want to find an equivalent of the rest : \begin{equation} R_n=\sum_{k=n+1}^{\infty}{\frac{1}{k^3}} \end{equation}

Answer 1

Summing the inequalities $$\frac1{k^2}-\frac1{(k+1)^2}\leqslant\frac2{k^3}\leqslant\frac1{(k-1)^2}-\frac1{k^2}$$ yields $$\frac1{2(n+1)^2}\leqslant R_n\leqslant\frac1{2n^2}.$$ From here, one sees that the RHS is a simple equivalent of $R_n$ when $n\to\infty$.

More generally, for every $\alpha\gt0$, $$\sum_{k=n+1}^\infty\frac1{k^{1+\alpha}}\sim\int_n^\infty\frac{\mathrm dx}{x^{1+\alpha}}=\frac1{\alpha n^\alpha},$$ where the equivalent sign means that the ratio of the two sides of the sign converges to $1$. "Hard" inequalities, similar to the ones above and using only the fact that the function $x\mapsto 1/x^{1+\alpha}$ is nonincreasing and is a multiple of the derivative of $x\mapsto 1/x^{\alpha}$, are $$\frac1{\alpha (n+1)^\alpha}=\int_{n+1}^\infty\frac{\mathrm dx}{x^{1+\alpha}}\leqslant\sum_{k=n+1}^\infty\frac1{k^{1+\alpha}}\leqslant\int_n^\infty\frac{\mathrm dx}{x^{1+\alpha}}=\frac1{\alpha n^\alpha}.$$

Answer 2

Use the theorem: if $x_n \sim y_n > 0$ and $\sum x_n < \infty$ then $$\sum_{n=N}^\infty x_n \sim \sum_{n=N}^\infty y_n $$

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Let $$u_n= \frac1{2n^2}$$

Then check that $u_n - u_{n+1}\sim \frac1{n^3}$ (you can find such a relation looking for a $u_n$ in the form $u_n = \frac a{x_m}$).

Hence $$ \sum_{n=N}^\infty \frac 1{n^3}\sim u_N = \frac 1{2N^2} $$