# How do you use the Nth term test on the infinite series sum_(n=1)^oorootn(2) ?

Recall that the ${n}^{t h}$ term test is a test of divergence only. It states that if the sequence of general terms ${\left\{{a}_{n}\right\}}_{n = 1}^{\setminus} \infty$ does not converge to 0, then the series $\setminus {\sum}_{n = 1}^{\setminus} \infty {a}_{n}$ is divergent. ATTN: If the limit is 0, nothing can be concluded from this test, and another test needs to be used to decide whether the series converges or not.
In the present case, ${a}_{n} = \setminus \sqrt[n]{2} = {2}^{\frac{1}{n}}$.
Since $\setminus {\lim}_{n \setminus \to \setminus \infty} {2}^{\frac{1}{n}} = {2}^{0} = 1 \setminus \ne 0$, we conclude by the ${n}^{t h}$ term test that the series $\setminus {\sum}_{n = 1}^{\setminus} \infty \setminus \sqrt[n]{2}$ is divergent.