Radius and interval of convergence Determine the radius and in...

Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₂∞ ((x+3)ᵏ)/(k łn²k)

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the radius and interval of convergence for the power series. The sum evaluated from N equals 3 to positive infinity of parentheses X minus 2 to the power of N divided by N multiplied by the natural log to the power of 3 of N. Awesome. So it appears for this particular prom we're asked to take this particular power series that is provided to us by the prom itself, and we're asked to solve for two separate answers. We're asked to solve for first, the radius of convergence, and secondly, we're asked to solve for the interval of convergence. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to recall and note that for this particular prompt, the general term will be a subscript N, which we'll call AN for short, is equal to parentheses X minus 2 to the power of N divided by N multiplied by the natural log to the power of 3 of N. And this is for N is greater than or equal to 3. Fantastic. Our second step now is we need to recall and apply the ratio test, so we need to write that the limit as N approaches positive infinity of the absolute value of A subscript N + 1 divided by a subscript N. Fantastic. Our third step is we need to substitute our values into our expression that we're using for the ratio test, and when we do, we will find that the absolute value we're going to be writing the absolute value of parentheses X minus 2 in parenthesis to the power of N + 1. Divided by parentheses, N + 1 multiplied by the natural log to the power of 3 of N + 1. Multiplied by N multiplied by the natural log to the power of 3 of N divided by parentheses X minus 2 to the power of N, fantastic. So, our 4th step is we need to simplify. And when we do, we will find that the absolute value of X minus 2 multiplied by N divided by N + 1 multiplied by the natural law to the power of 3 of N divided by the natural log to the power of 3 of N + 1. Fantastic. So now, moving right along here, our 5th step is we need to take the limit as N approaches positive infinity. So that will mean that N divided by N + 1 will approach 1, and that will mean that the natural log to the power of 3 of N divided by the natural log to the power of 3 of N + 1. We'll also approach positive one. So that will mean. That our limit is the absolute value of X minus 2. Fantastic. So moving right along here, our 6th step will be that as we should recall, by the ratio test, the series converges when the absolute value of X minus 2 is less than 1. So that means it will converge if this is the case. So I put C in quotation marks to mean converge. So that will mean our 7th step is that our radius of convergence R will be equal to 1. And that is one of our final answers. Hooray, we did it. So now, we need to do our 8th step, which is we need to recall and note that for the interval, the absolute value of X minus 2, so X minus 2 and absolute value signs is less than 1. So that will mean that 1 is less than X and X is less than 3. Fantastic. So for our 9th step, we need to check our end points. So we need to note that for our endpoints, X is equal to 1 and X is equal to 3. So, our step 10, focusing on X is equal to 1, that will mean that a subscript N is equal to parentheses -1 to the power of N divided by N multiplied by the natural log to the power of 3 of N, and we need to recognize that this is an alternating series. And that will mean that the term, so let's make a note that the terms 1 divided by N multiplied by the natural log to the power of 3 of N, decreases. So I'll put an arrow pointing downwards for decreases, and it will also approach 0. So it's decreasing and it approaches 0. Fantastic. So, since the natural log of 3 is greater than 1. Or N is greater than or equal to 3, that will mean that we have the natural log to the power of 3 of 3 is greater than 1, and that N multiplied by the natural log of 3, sorry, that N multiplied by the natural log to the power of 3 of 3 is greater than 1, as N increases. The denominator will also increase. Fantastic. So that means that it will result in decreasing terms, so that, as we should recall at this point, by the alternating series test, that means that the series will also Converge, fantastic. So our step 11 is we need to note that for X is equal to 3, the series will become the sum evaluated from N is equal to 3 to positive infinity of 1 divided by N multiplied by the natural log to the power of 3 of N. So when we perform the integral test, we need to recall that we need to consider the following interval for this particular prom, that the integral evaluated from 3 to positive infinity of 1 divided by X multiplied by the natural log to the power 3 of XDX, and for this particular interval. As integral in order to evaluate it, we need to let U equal the natural log of X. So that will mean that DU is equal to 1 divided by XDX. So that will mean that we can go ahead and write that the integral evaluated from 3 to positive infinity of 1 divided by X multiplied by the natural log to the power of 3 of X D X is equal to the integral evaluated from the natural log of 3 to positive infinity of 1 divided by U to the power of 3DU. Which will be equal to the limit, as B approaches positive infinity of square brackets of negative, 1 divided by 2 multiplied by U to the power of 2, evaluated from the natural log of 3 to B. Which will be equal to the limit as B approaches positive infinity of square brackets of -1 divided by 2 multiplied by B to the 2, plus 1 divided by 2 multiplied by the natural log to the power of 2 of 3, fantastic. So, that will mean that as B approaches positive infinity, that will mean that -1 divided by 2 multiplied by B to the 2 also approaches 0, so that means that the integral converges. So, therefore, as a result, the series, the sum evaluated from N is equal to 3 to positive infinity of 1 divided by N multiplied by the natural log to the power of 3 of N converges at X equals 3, according to the interval test. Fantastic. So our 12th and final step is we need to create a summary for our endpoints. So as we should recall and note that since the series converges at both endpoints, we will need to include them in the interval of convergence. So that will mean that without a doubt, the final interval of convergence will be square. Brackets of 1.3, where the square brackets indicate a closed interval. And even though we don't, we're not dealing with parentheses, just recall that parentheses mean an open interval, whereas a square bracket indicates a closed interval. And that's it. Those are our final answers. Hooray, we did it. So going back to the top to recap what we've learned, that will mean that our final two answers, without a doubt, will have to be once again the following. So our radius of convergence will be R is equal to 1 and our interval of convergence will be square brackets of 1.3. And that's it. Those are our final two answers. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₂∞ ((x+3)ᵏ)/(k łn²k)

Key Concepts

Radius of Convergence

Interval of Convergence

Ratio Test for Convergence

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