Question Video: Using the Root Test to Determine Convergence | Nagwa Question Video: Using the Root Test to Determine Convergence | Nagwa

Question Video: Using the Root Test to Determine Convergence

Consider the series β_(π = 0)^(β) π_π, where π_π = (π + 1)^(π)/6^(2π). Calculate lim_(π β β) |π_π|^(1/π). Hence, determine whether the series converges or diverges.

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Video Transcript

Consider the series the sum from π equal zero to β of π sub π, where π sub π is equal to π plus one to the πth power over six to the power of two π. Calculate the limit as π approaches β of the absolute value of π sub π to the power of one over π. And hence, determine whether the series converges or diverges.

We have had π sub π defined for us. Itβs π plus one to the πth power over six to the power of two π. And so, to answer the first part of this question, we need to establish the limit as π approaches β of the absolute value of π plus one to the πth power over six to the power of two π all to the power of one over π.

Now, in fact, to make this a little bit easier, weβre going to rewrite the part inside the absolute values signs. Weβre going to write it as π plus one over six squared all to the πth power. Then, we can take that πth power outside the absolute value signs. And we have the limit as π approaches β of the absolute value of π plus one over six squared to the πth power all to the power of one over π.

Now, when we have the brackets, we multiply the exponents. And π times one over π is simply one. And so, weβre looking to find the limit as π approaches β of the absolute value of π plus one over six squared. Now, of course, one over six squared is independent of π. So, we can take that outside of our limit. And now, we can see weβre looking to compute one over six squared times the limit as π approaches β of the absolute value of π plus one.

One over six squared is one over 36. But as π gets larger, the absolute value of π plus one gets larger. In other words, it also approaches β. So, we have one over 36 times β, which is also β. And so, we calculated the limit as π approaches β of the absolute value of π sub π to the power of one over π. Itβs β.

The second part of this question asks us to determine whether the series converges or diverges. Well, what weβve actually set up for is to use the root test. Letβs call the limit as π approaches β of the absolute value of π sub π, all to the power of one over π, π. If this value of π is less than one, then our series, the sum of π sub π, is absolutely convergent and, therefore, convergent. If π is greater than one, the series diverges. And actually, if itβs equal to one, the root test is inconclusive. Well, β is quite clearly greater than one. And so, we deduce that the series diverges.

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