Question Video: Using the Limit Comparison Test Where the Summant Is a Reciprocal of the Sum of Exponential Functions | Nagwa Question Video: Using the Limit Comparison Test Where the Summant Is a Reciprocal of the Sum of Exponential Functions | Nagwa

Question Video: Using the Limit Comparison Test Where the Summant Is a Reciprocal of the Sum of Exponential Functions Mathematics • Higher Education

Use the limit comparison test to determine whether the series ∑_(𝑛 = 1)^(∞) 1/(2^(𝑛) + 3𝑛) is convergent or divergent.

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Video Transcript

Use the limit comparison test to determine whether the series the sum from 𝑛 equals one to ∞ of one divided by two to the 𝑛th power plus three 𝑛 is convergent or divergent.

The question wants us to use the limit comparison test to determine the convergence or divergence of our series. And we recall the limit comparison test tells us for two series, the sum of 𝑎 𝑛 and the sum of 𝑏 𝑛, where both 𝑎 𝑛 and 𝑏 𝑛 are positive for all values of 𝑛. If the limit as 𝑛 approaches ∞ of the quotient 𝑎 𝑛 divided by 𝑏 𝑛 is equal to some positive finite value 𝑐, then we can conclude that either both of our series are convergent or both of our series are divergent.

By looking at the summand in our series, we can see that it’s very similar to the geometric series, one divided by two to the 𝑛th power. So, in our limit comparison test, we’ll set the sequence 𝑏 𝑛 to be equal to the summand of the series given to us in the question. That’s one divided by two to the 𝑛th power plus three 𝑛. And we’ll set our sequence 𝑎 𝑛 to be equal to the geometric sequence one divided by two to the 𝑛th power.

To use our limit comparison test, we need both 𝑎 𝑛 and 𝑏 𝑛 to be positive for all values of 𝑛. We see from our series the values of 𝑛 start when 𝑛 is equal to one. When 𝑛 is greater than or equal to one, three 𝑛 will be positive and two to the 𝑛th power will also be positive. Therefore, their sum is positive. And we know one divided by a positive number is positive. Therefore, our sequence 𝑏 𝑛 is a positive sequence for all values of 𝑛. Similarly, we see that two to the 𝑛th power is positive for all values of 𝑛 and one divided by two to the 𝑛th power is also positive for all values of 𝑛. So, 𝑎 𝑛 is positive for all values of 𝑛.

So, because both of our sequences are positive for all values of 𝑛, we can try to use the limit comparison test. We have the limit as 𝑛 approaches ∞ of 𝑎 𝑛 divided by 𝑏 𝑛 is equal to the limit as 𝑛 approaches ∞ of one divided by two to the 𝑛th power multiplied by the reciprocal of 𝑏 𝑛, which is two to the 𝑛th power plus three 𝑛. We can simplify this by dividing through by our denominator of two to the 𝑛th power.

Dividing through by two to the 𝑛th power gives us the limit as 𝑛 approaches ∞ of one divided by three 𝑛 divided by two to the 𝑛th power. And we can see as 𝑛 is approaching ∞, our numerator of three 𝑛 is approaching ∞ and our denominator of two to the 𝑛th power is also approaching ∞, which gives us an indeterminate form. However, we can try to evaluate the quotient of two functions by using L’Hôpital’s rule.

We recall one version of L’Hôpital’s rule states that if functions 𝑓 and 𝑔 are differentiable where the derivative of 𝑔 is never equal to zero and both the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 and the limit as 𝑥 approaches ∞ of 𝑔 of 𝑥 are both equal to ∞. Then if the limit as 𝑥 approaches ∞ of 𝑓 prime of 𝑥 divided by 𝑔 of 𝑥 exists or is equal to positive or negative ∞. Then we must have the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 divided by 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches ∞ of 𝑓 prime of 𝑥 divided by 𝑔 prime of 𝑥.

So to apply L’Hôpital’s rule, we’ll set our function 𝑓 to be the function in the numerator and our function 𝑔 to be the function in the denominator. That’s 𝑓 of 𝑛 is three 𝑛 and 𝑔 of 𝑛 is two to the 𝑛th power. The first prerequisite to using L’Hôpital’s rule is that both 𝑓 and 𝑔 are differentiable. We see that 𝑓 is a polynomial, and so it is differentiable. And we see that 𝑔 is an exponential function, and so it’s also differentiable.

Next, we need to show that our derivative function 𝑔 is never equal to zero. We can show this by rewriting our function 𝑔 of 𝑛 as 𝑒 to the power of the natural logarithm of two to the 𝑛th of power. Then, using our power rule for logarithms, we have the natural logarithm of two to the 𝑛th power is equal to 𝑛 times the natural logarithm of two.

Now, to differentiate our function 𝑔 of 𝑛, we’re differentiating 𝑒 to the power of 𝑛 times the natural logarithm of two. And we know for any constant 𝑎, the derivative of 𝑒 to the power of 𝑎𝑥 with respect to 𝑥 is just equal to 𝑎 times 𝑒 to the power of 𝑎𝑥. So, 𝑔 prime of 𝑛 is equal to the natural log of two times 𝑒 to the power of 𝑛 times the natural log of two. And we know for any value of 𝑛, the exponential function will be positive and the natural logarithm of two is also positive. So, we’re multiplying two positive numbers together, giving us a positive number. So, our derivative function is positive and therefore it’s not equal to zero.

We chose this version of L’Hôpital’s rule because our quotient was indeterminant of the form ∞ divided by ∞. So, both of these limits are equal to ∞. Now, we need to calculate the limit as 𝑛 approaches ∞ of 𝑓 prime of 𝑛 divided by 𝑔 prime of 𝑛. So, let’s calculate this limit. If 𝑓 of 𝑛 is equal to three 𝑛, then the derivative is equal to three. This gives us the limit as 𝑛 approaches ∞ of 𝑓 prime of 𝑛 divided by 𝑔 prime of 𝑛 is equal to the limit as 𝑛 approaches ∞ of three divided by the natural log of two times 𝑒 to the power of 𝑛 times the natural log of two. And we can evaluate this limit. As 𝑛 is approaching ∞, both three and the natural logarithm of two remain constant. However, the exponential function in the denominator approaches ∞. Therefore, our limit is approaching zero.

So, we’ve shown that this limit exists. And since this is equal to zero, by L’Hôpital’s rule, we can conclude the limit as 𝑛 approaches ∞ of 𝑓 of 𝑛 divided by 𝑔 of 𝑛 must be equal to zero. So, going back to the limit as 𝑛 approaches ∞ of one plus three 𝑛 divided by two to the 𝑛th power, we have the limit of a sum is equal to the sum of the limits. And the limit of the constant one is just equal to one. This gives us one plus the limit as 𝑛 approaches ∞ of three 𝑛 divided by two to the 𝑛th power. And we showed by using L’Hôpital’s rule that this limit is equal to zero. So, we must have that this is just equal to one.

Therefore, we’ve shown the limit as 𝑛 approaches ∞ of 𝑎𝑛 divided by 𝑏𝑛 is equal to one, which is a finite, positive constant. So, we’ve shown all the prerequisites of our limit comparison test are true. So, by using the limit comparison test, our infinite series the sum of 𝑎 𝑛 and the sum of 𝑏 𝑛 must either both converge or they must both diverge. And we chose our sequence 𝑎 𝑛 to be a geometric sequence with ratio of successive terms equal to one-half.

And we know the infinite sum of a geometric series will always converge if the absolute value of the ratio 𝑟 is less thsan one. Using this, we have that the infinite sum of our geometric sequence 𝑎 𝑛 must be convergent since it’s a geometric series with ratio of successive terms one-half. And the limit comparison test told us that either both of our series were convergent or both of our series were divergent. Therefore, we have the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 must also be convergent.

In conclusion, we’ve shown by using the limit comparison test the sum from 𝑛 equals one to ∞ of one divided by two to the 𝑛th power plus three 𝑛 must be convergent.

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