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

Question

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

Ezekiel Alexander · Accepted Answer

Recall that the #n^{th}# term test is a test of divergence only. It states that if the sequence of general terms #{a_n}_{n=1}^\infty# does not converge to 0, then the series #\sum_{n=1}^\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=oot[n]2=2^{1/n}#. 
Since #\lim_{n	o\infty} 2^{1/n}=2^0=1
e 0#, we conclude by the #n^{th}# term test that the series #\sum_{n=1}^\infty oot[n]2# is divergent.

Samantha Adkison · Answer

undefined To use the Nth term test on the infinite series $ \sum_{n=1}^{\infty} \sqrt{n}(2) $, follow these steps:

1. Evaluate the Nth term of the series, which is $ \sqrt{n}(2) $.
2. Determine the behavior of the Nth term as $ n $ approaches infinity.
3. Apply the Nth term test, which states that if the limit of the Nth term as $ n $ approaches infinity does not equal zero, then the series diverges.
4. If the limit is zero, the test is inconclusive, and other convergence tests may need to be applied.

Therefore, for the series $ \sum_{n=1}^{\infty} \sqrt{n}(2) $, calculate the limit of $ \sqrt{n}(2) $ as $ n $ approaches infinity. If the limit is not zero, the series diverges; if the limit is zero, further tests may be necessary to determine convergence.