# Question Video: Using the Ratio Test Mathematics • Higher Education

True or false: The series β_(π = 1)^(β) (1/(πΒ² + 1)) is convergent by the ratio test.

02:22

### Video Transcript

True or false: The series, which is the sum from π equals one to β of one over π squared plus one, is convergent by the ratio test.

Weβll need to apply the ratio test to this series. The ratio test tells us that for the series, which is the sum from π equals one to β of π π, where πΏ is equal to the limit as π tends to β of the absolute value of π π plus one over π π. Then, firstly, if πΏ is less than one, then the series converges absolutely. Secondly, if πΏ is greater than one, then the series diverges. And thirdly, if πΏ is equal to one, then the test is inconclusive.

From the series given in the question, we have that π π is equal to one over π squared plus one. Therefore, π π plus one is equal to one over π plus one squared plus one, which after we distribute the parentheses, we can see is equal to one over π squared plus two π plus two. Now, weβre ready to find our value of πΏ. We have that πΏ is equal to the limit as π tends to β of π π plus one over π π. Substituting in our values for π π and π π plus one, we can see that this limit is equal to the limit as π tends to β of the absolute value of π squared plus one over π squared plus two π plus two.

Now, we can divide both the numerator and denominator of the fraction here by π squared. And weβll do this because weβre trying to find the infinite limit of a rational function. And π squared is the highest power which occurs in our fraction. So we have the limit as π tends to β of the absolute value of one plus one over π squared all over one plus two over π plus two over π squared. Next, we used the fact that the limit as π tends to β of one over π is equal to zero. Therefore, any term within our limit, which uses one over π or one over π squared, will tend to zero as π tends to β.

And so our limit will become the absolute value of one plus zero over one plus zero plus zero. And this is simply equal to one. Therefore, we found that πΏ is equal to one. Looking at the ratio test, we can see that this satisfies condition number three. Therefore, we can say that the ratio test is inconclusive. It doesnβt tell us whether this series is absolutely convergent, conditionally convergent, or divergent. Therefore, the answer to this question is false.