# Question Video: Determine the Type of Convergence for an Alternating Series Mathematics • Higher Education

State whether the series ∑_(𝑛 = 1) ^(∞) (−1)^(𝑛 + 1) 𝑛^(𝑛) converges absolutely, conditionally, or not at all.

02:20

### Video Transcript

State whether the series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times 𝑛 to the 𝑛th power converges absolutely, conditionally, or not at all.

We’re given a series and we’re asked to determine whether this series converges absolutely, conditionally, or whether the series does not converge at all. And at this point, we have a lot of different methods for dealing with the convergence or divergence of different series. And it can often be hard to decide which one we should proceed with. In this case, though, we can see something interesting about our summands. First, we can see we have negative one to the power of 𝑛 plus one. This is just going to alternate the sign of each of our terms. And we can also see we multiply this by 𝑛 to the 𝑛th power, which we know is going to grow without bound. So the size of each term in our series is growing without bound.

And if the terms of our series are not approaching zero, this should remind us of the 𝑛th-term divergence test. The 𝑛th-term divergence test tells us that the terms in our series must approach zero for our series to converge. In other words, if the limit as 𝑛 approaches ∞ of 𝑎 𝑛 is not equal to zero or the limit as 𝑛 approaches ∞ of 𝑎 𝑛 does not exist, then the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 must be divergent. In our case, our summand 𝑎 𝑛 is negative one to the power of 𝑛 plus one times 𝑛 to the 𝑛th power. So we can look at the limit as 𝑛 approaches ∞ of negative one to the power of 𝑛 plus one times 𝑛 to the 𝑛th power.

Of course, we already know what happens as 𝑛 approaches ∞. We’ve already said negative one to the power of 𝑛 plus one just changes the sign of each term. And 𝑛 to the 𝑛th power is growing without bound. And if the size of each term is growing without bound, then in particular our limit does not exist. And it’s worth reiterating here being equal to positive ∞ or being equal to negative ∞ or oscillating between multiple values are examples of a limit not existing. So by the 𝑛th-term divergence test because the limit as an approaches ∞ of our summand is not equal to zero, our series must be divergent.

And it’s also worth pointing out if we were to take the absolute value of each term in our series, we would end up with the same conclusion. The only difference is our limit would be the limit as 𝑛 approaches ∞ of just 𝑛 to the 𝑛th power, which grows without bound. And so this limit does not exist. But in either way, we can use the 𝑛th-term divergence test to show that our series must be divergent. Therefore, we were able to show the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times 𝑛 to the 𝑛th power does not converge at all.