Consider the following infinite sum and its partial sums.
$$ S = \sum_{k=1}^{\infty } \frac{1}{k^3} $$
$$ S_N = \sum _{k=1}^{N} \frac{1}{k^3} $$
Use integral comparison to estimate how fast $S_N$ tends to $S$ . That is, show that $0<S−S_N<CN^{−q}$ giving the best values you can for the constants $C$ and $q$ .
My work is as follows:
The lower Riemann sum is given by $\sum_{k=2}^{\infty} \frac{1}{k^3}$. The upper Riemann sum is $\sum_{k=1}^{\infty} \frac{1}{k^3}$.
Since $ S = \sum_{k=1}^{\infty } \frac{1}{k^3} $, the the lower Riemman Sum is $S-1 = \sum_{k=2}^{\infty} \frac{1}{k^3}$.
Thus, $S-1 < \int_{1}^{\infty} \frac{dx}{x^3} < S$. We could evaluate the improper integral if it converges, however it shows to be inconclusive. It happens to be equal to $\zeta(3)$, however that is out of scope of the class I am taking and it is not accepted.
I don't know where to go from there. Also, what can I do with $ S_N = \sum _{k=1}^{N} \frac{1}{k^3} $, I am stuck on how to proceed.
Hint: $$\int_{N+1}^\infty \frac{1}{x^3}\; dx < S - S_N = \sum_{n=N+1}^\infty \frac{1}{n^3} < \int_{N}^\infty \frac{1}{x^3}\; dx $$ In fact, you can do slightly better: since $1/x^3$ is convex for $x > 0$, $$\int_{N+1/2}^\infty \frac{1}{x^3}\; dx \ge \sum_{n=N+1}^\infty \frac{1}{n^3} $$