Is $\lim_{m\to\infty}\sum_{k≥t}\frac{1}{\binom kt^m}=1$?

Question

Context:

Using Wolfram calculator, I've observed that :

$$\sum_{k≥2}\frac{1}{\binom k2^{100}}≈1$$

$$\sum_{k≥5}\frac{1}{\binom k5^{100}}≈1$$

$$\sum_{k≥4}\frac{1}{\binom k4^{50}}≈1$$

Question:

I want to know if it's true that

$$\lim_{m\to\infty}\sum_{k≥t}\frac{1}{\binom kt^m}=1$$

.

Also, how can we proceed to prove it for general case ? Unfortunately I can't attempt to prove because my knowledge related to higher power sums is very low. Thanks !

Answer 1

For any $t\geq 2, m\geq 2,$

$$\sum_{k=t}^{\infty}\frac{1}{{\binom kt}^{m}} \leq \sum_{k=1}^{\infty}\frac{1}{k^{m}} \leq 1 + \int_{x=1}^{\infty} \frac{1}{x^m} dx = 1 + \left[ - \frac{1}{m-1} x^{-(m-1)} \right]_{x=1}^{\infty} = 1 + \frac{1}{m-1} \to 1 \text{ as } m \to \infty. $$