Limit of a series

Calculus Level pending

Calculate the following:

$\displaystyle \large \lim_{n \to \infty} \sum_{k=0}^{n} \dfrac{1}{\sqrt{16n+3k^{3}}}$

The answer is 0.

1 solution

Brian Moehring
Dec 2, 2018

Note that for every pair $n>0, k>0,$ $0 \leq \frac{1}{\sqrt{16n+3k^3}} \leq \frac{1}{k^{3/2}}$ and also that $\sum_{k=1}^\infty \frac{1}{k^{3/2}}$ converges by the p-series test [alternatively, the integral test].

Therefore, by the Dominated Convergence Theorem, $\lim_{n\to\infty} \sum_{k=0}^n \frac{1}{\sqrt{16n+3k^3}} = \sum_{k=0}^\infty \lim_{n\to\infty} \frac{1}{\sqrt{16n+3k^3}} = \sum_{k=0}^\infty 0 = \boxed{0}$

I tried to find a more elementary proof initially, but found an embarrassing algebra mistake in my attempt I couldn't surmount using the same methods.

I would be interested to see any truly elementary proof [read: any proof that doesn't use results from measure theory]

Brian Moehring - 2 years, 6 months ago

Limit of a series

The answer is 0.

1 solution

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