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When working on series and integrals, I attained two interesting limits: $$ \lim_{M\to \infty}\left\{\sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M\over \alpha+1}\right\}=-{1\over 2} $$ $$ \lim_{M\to \infty}\left\{M\cdot \sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M^2\over \alpha+1}+{M\over 2}\right\}=-{\alpha\over 12} $$ for $\alpha>0$. Now my question is:

How could these limits be found using the definition of integral? One I my friends advised to use a trapezoid approach for that, but I'm suspicious. Would it give me the answer to both of the limits?

Also

what is the value for the following limit? $$ \lim_{M\to \infty}\left\{M^2\cdot \sum_{i=0}^{M-1}\left({i\over M}\right)^\alpha-{M^3\over \alpha+1}+{M^2\over 2}+{M\alpha\over 12}\right\} $$

Mostafa Ayaz
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2 Answers2

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$$\frac 1M \sum_{i=0}^{M-1}\left(\frac iM\right)^\alpha$$ is an approximation for the integral $$ \int_0^1 x^\alpha \, dx = \frac{1}{\alpha+1} \, . $$

The difference can be computed with Taylor's formula for $f(x) = x^{\alpha+1}$ on each interval $[\frac iM, \frac{i+1}{M}]$: $$ \int_{i/M}^{(i+1)/M} x^\alpha \, dx = \frac{1}{\alpha+1} \left(\left(\frac {i+1}M\right)^{\alpha+1} - \left(\frac iM\right)^{\alpha+1} \right) \\ = \frac 1M \left(\frac iM\right)^\alpha + \frac{\alpha}{2M^2} \left(\frac iM\right)^{\alpha-1} + O\left( \frac{1}{M^3}\right) \, . $$ Now sum these identities and multiply by $M$. It follows that $$ \frac{M}{\alpha+1} - \sum_{i=0}^{M-1}\left(\frac iM\right)^\alpha = \frac{\alpha}{2} \frac 1M \sum_{i=0}^{M-1}\left(\frac iM\right)^{\alpha-1} + O\left( \frac{1}{M}\right) $$

For $M \to \infty$ the right-hand side has the limit $$ \frac{\alpha}{2} \int_0^1 x^{\alpha -1} \, dx = \frac 12 \, , $$ that is your first limit.

The second and third limit can be obtained in the same way, by using more terms in Taylor's formula for $f(x) = x^{\alpha+1}$.

Martin R
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If $\alpha$ is a positive integer then your limits can be easily derived from the Faulhaber's formula

$$\sum _{i=0}^{M-1}i^{\alpha}=\frac{1}{\alpha+1}\sum _{j=0}^{\alpha}{\alpha+1 \choose j}B_{j}(M-1)^{\alpha+1-j}$$ where $B_j$ is the $j$-th Bernoulli number.

Robert Z
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  • I am just curious: Does that require $\alpha$ to be an integer, or does it work with fractional exponents as well? – Martin R Aug 02 '19 at 13:10
  • Thank you for your answer. As @MartinR said and hence editing my question, does it also work for any $\alpha>0$? – Mostafa Ayaz Aug 02 '19 at 13:18
  • I assumed that $\alpha$ is a positive integer. See https://math.stackexchange.com/questions/82588/is-there-a-formula-for-sums-of-consecutive-powers-where-the-powers-are-non-inte – Robert Z Aug 02 '19 at 13:29