## Jul 23, 2018

It is converging.

#### Explanation:

To figure out if a sum of a series of terms converges or diverges we need to decide if the terms are getting bigger or smaller.

In this case the terms are summed for values of $k = 1$ to $k = \infty$.

We need to work out if the value of the expression increases as $k$ increases through this range of values.

Looking at the expression we can see that the numerator $\sqrt{{k}^{2} + 1}$ is a square root, which has value

$\sqrt{2} = 1.414 \left(3 \mathrm{dp}\right)$ at $k = 1$.

It is then going to get closer and closer to k as k increases, so it is increasing at very nearly the same rate as k once k is large.

e.g. at $k = 10 , \left(\sqrt{{k}^{2} + 1}\right) = \sqrt{101} = 10.050 \left(3 \mathrm{dp}\right)$

which is already pretty close to $k$

The denominator is 3 less than ${k}^{3}$, starting at -2 when $k = 1$ and increasing at close to ${k}^{3}$ as $k$ increases.

e.g. at $k = 10 , \left({k}^{3} - 3\right) = 997$

So we see that as $k \to \infty$, the expression $\frac{\sqrt{{k}^{2} + 1}}{{k}^{3} - 3} \to \frac{k}{k} ^ 3 \to \frac{1}{\infty}$

So it is clearly converging.