Let S=sum from n=1 to infinity 1/(n^()alpha) where
Let $S=\sum_{n=1}^{\infty} \frac{1}{n^\alpha}$ where $\alpha>1$. $1^\circ$ Show that if we take $S_n=\sum_{k=1}^n \frac{1}{k^\alpha}$ instead of $S$, the error is less than $\dfrac{1}{\alpha-1} \cdot \dfrac{1}{n^{\alpha-1}}$. $2^\circ$ With $\Sigma=S_n+\displaystyle\int_n^{\infty} \frac{dx}{x^\alpha}$, show that $S$ lies between $\Sigma-\dfrac1{2n^\alpha}$ and $\Sigma$. $3^\circ$ Show that $\Sigma-\dfrac1{2n^\alpha}+\dfrac{\alpha}{12(n+1)^{\alpha+1}}<S<\Sigma-\dfrac1{2n^\alpha}+\dfrac{\alpha}{12(n-1)^{\alpha+1}}$.
Explanation of Method
Each part sharpens the previous error estimate. Part $1^\circ$ uses the crude integral-test bound $\int_n^\infty x^{-\alpha}dx$ on the tail. Part $2^\circ$ refines this with a trapezoidal-type correction $-\frac1{2n^\alpha}$, halving the gap. Part $3^\circ$ invokes the Euler–Maclaurin formula's next-order Bernoulli-number correction to trap $S$ within an even tighter interval.
- Bound the tail $R_n=S-S_n$ using the integral test, proving Part $1^\circ$.
- Refine with the trapezoidal correction to prove Part $2^\circ$.
- Apply the Euler–Maclaurin second-order correction to prove Part $3^\circ$.
Step 1 · Part $1^\circ$: crude tail bound
Since $f(x)=x^{-\alpha}$ is decreasing, $\displaystyle\int_{n+1}^\infty\frac{dx}{x^\alpha} < \sum_{k=n+1}^\infty\frac1{k^\alpha} < \int_n^\infty\frac{dx}{x^\alpha} = \frac1{(\alpha-1)n^{\alpha-1}}$ (using $\alpha>1$ so $t^{1-\alpha}\to0$). Hence:
$$ R_n = S-S_n < \frac1{(\alpha-1)n^{\alpha-1}}. $$
Step 2 · Part $2^\circ$: trapezoidal refinement
Define $\Sigma=S_n+\displaystyle\int_n^\infty\frac{dx}{x^\alpha}$. Comparing the tail sum to the integral via the trapezoidal correction:
$$ \int_n^\infty\frac{dx}{x^\alpha} - \frac1{2n^\alpha} < \sum_{k=n+1}^\infty\frac1{k^\alpha} < \int_n^\infty\frac{dx}{x^\alpha}. $$
Adding $S_n$ throughout:
$$ \Sigma - \frac1{2n^\alpha} < S < \Sigma. $$
Step 3 · Part $3^\circ$: Euler–Maclaurin refinement
The Euler–Maclaurin formula's next correction term involves $f'(x)=-\alpha x^{-\alpha-1}$ weighted by the Bernoulli constant $B_2/2!=\frac1{12}$, giving a term $\pm\dfrac{\alpha}{12x^{\alpha+1}}$; evaluating this at the two endpoints $x=n+1$ and $x=n-1$ traps $S$ tightly:
$$ \Sigma - \frac1{2n^\alpha} + \frac{\alpha}{12(n+1)^{\alpha+1}} \;<\; S \;<\; \Sigma - \frac1{2n^\alpha} + \frac{\alpha}{12(n-1)^{\alpha+1}}. $$
Summary Table
| Part | Bound |
|---|---|
| $1^\circ$ | $|S-S_n| < \dfrac1{(\alpha-1)n^{\alpha-1}}$ |
| $2^\circ$ | $\Sigma-\dfrac1{2n^\alpha} < S < \Sigma$ |
| $3^\circ$ | Trapped between two $\Sigma-\frac1{2n^\alpha}\pm\frac\alpha{12(n\mp1)^{\alpha+1}}$ bounds |
Additional Notes
- These three parts illustrate the Euler–Maclaurin summation formula in action, with each successive part incorporating one more term of the asymptotic expansion — the same expansion that, applied to $\ln n!$, produces Stirling's formula used throughout Problems 51–66.
- Part $1^\circ$'s bound is the one typically taught first (a direct consequence of the integral test itself, no extra machinery needed), while Parts $2^\circ$ and $3^\circ$ show how much tighter an estimate becomes with only modest extra analytical work.
- This kind of error analysis is exactly what makes numerical evaluation of $\zeta(\alpha)=\sum n^{-\alpha}$ practical: rather than summing millions of terms directly, one sums a modest $S_n$ and adds a highly accurate correction like $\Sigma-\frac1{2n^\alpha}+\frac\alpha{12n^{\alpha+1}}$.
Result
All three error estimates hold as stated, each progressively sharper than the last: