Answer
Converges
Work Step by Step
The kth term formula of the series is given by: $a_k=(\dfrac{k^2}{2k^2+1})^{k}$
Root Test states that when $\Sigma a_k$ is an infinite series with positive terms and, then $r=\lim\limits_{k \to \infty}\sqrt[k] a_k$
a) When $0 \leq r \lt 1$, the series converges. (b) When $r \gt 1$, or, $\infty$, so the series diverges. (c) When $r=1$, the ratio test is inconclusive.
Now, $r=\lim\limits_{k \to \infty}\sqrt[k] {(\dfrac{k^2}{2k^2+1})^{k}} \\= \lim\limits_{k \to \infty} \dfrac{k^2}{2k^2+1} \\=\dfrac{1}{2}$
Therefore, the series converges by the root test.
