MathJax Test

Posted on Wed 29 April 2020 in Posts

This is a test to see if MathJax works on blogger.

For \(\alpha, \beta \gt -1\), find the value of \(\lim_{n \to \infty} n^{\beta - \alpha} \frac{1^\alpha + 2^\alpha + 3^\alpha + ... + n^\alpha}{1^\beta + 2^\beta + 3^\beta + ... + n^\beta}\)

So we have

$$L = \lim_{n \to \infty} \frac{n^\beta}{n^\alpha} \times \frac{\sum_{r=0}^n r^\alpha}{\sum_{r=0}^n r^\beta}\ = \lim_{n \to \infty} \frac{\sum_{r=0}^n (\frac rn)^\alpha}{\sum_{r=0}^n (\frac rn)^\beta}$$

The summations can now be evaluated as Riemann sums by multiplying numerator and denominator by (\frac 1n)

$$L = \frac{\int_0^1 x^\alpha dx}{\int_0^1 x^\beta dx}\ L = \frac{\beta+1}{\alpha +1}$$

Indeed it does! Looking forward to writing more mathematical posts.