Answer
Diverges
Work Step by Step
Recall the property that the series $\Sigma_{n=1}^\infty a_n$ diverges if $\lim\limits_{n \to \infty} a_n \ne 0$.
Now, we have :
$\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{k}{\sqrt {k^2+1}}$
or, $=\lim\limits_{k \to \infty} \dfrac{k/k}{\sqrt {k^2+1}/k}$
or, $=\lim\limits_{k \to \infty} \dfrac{1}{\sqrt {1+1/k^2}}$
or, $=1$
So, we can see that $\lim\limits_{k \to \infty} a_k\ne 0$. Thus, the given series diverges.
