Evaluate the limits in Exercise 9 and 10 by identifying them w...

Question

Evaluate the limits in Exercise 9 and 10 by identifying them with definite integrals and evaluating the integrals.

lim (n → ∞) Σ (from k=1 to n) ln √(1 + k/n)

Accepted Answer

Hello there. Today we are going to 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. Evaluate the limit by interpreting it as a Riemann sum. And then computing the corresponding integral. The limit as n approaches infinity of the sum evaluated from k is equal to 1 ton of 1 divided by n multiplied by the natural log of 1 + 2 k divided by n. OK. So it appears for this particular prompt we're ultimately asked to evaluate the limit that is provided to us by the prompt itself and we're asked to evaluate this limit by interpreting it as a Riemann sum and then we're asked to compute the corresponding integral. OK, so with that in mind. Our first step that we need to take is based on our prior knowledge that in order to interpret the sum as a Riemann sum, we will need to set X subscript K, which I'll just call XK for short, we will set XK. Is equal to a divided by n. And we need to note, based on our prior knowledge that delta X will be equal to 1 divided by N. So once again, we are taking these steps in order to interpret out the sum that is provided to us by the problem itself. We're using this sum to interpret it as a Riemann sum. And now, with this knowledge in mind, we can state that the sum evaluated from k is equal to 1 ton of 1 divided by n multiplied by the natural log of 1 + 2k divided by n. We can go ahead and write. That the integral evaluated from 0 to 1 of the natural log of 1 + 2 x dx, and this is as n approaches infinity, specifically as n approaches positive infinity. So, with that said, we now need to compute the integral by parts. So as we should recall, by using the integral by parts method, we need to let you be equal to the natural log of 1 + 2 X, and we need to let DV be equal to DX. So, therefore, we could go ahead and write that DU is equal to 2 divided by 1 + 2 X. Multiplied by DX and we need to note that V will be equal to X. So, as we should recall, so this will be our second step, the integration by parts method equation is written as, as we should recall, the integral of UDV is equal to u multiplied by V minus the integral of VDU. So, with that in mind, we can go ahead and write the integral of the natural log of 1 + 2 X. DX will be equal to x multiplied by the natural log of. 1 + 2 x minus the integral of. 2 X. Divided by one plus. 2 X. DX. OK. Moving right along here. So now, for our next step, so let's call this step 3, we can also note that we could write 2 X divided by 1 plus 2 X as drum roll, we could write this as. 2 x divided by 1 + 2 x is equal to 1 + 2 x divided by 1 + 2 x minus 1 divided by 1 + 2 x, which will be equal to 1 minus 1 divided by 1 + 2 x. So, therefore, drum roll, therefore, without a doubt, we can go ahead and write the integral. So therefore, we can write that the integral of the natural log of 1 + 2 x, the X is equal to x multiplied by the natural log of. 1 + 2 x minus the integral of 1 minus 1 divided by 1 + 2 x in parentheses dx. Which will be equal to x multiplied by the natural log of 1 + 2 x minus x minus 1/2 multiplied by the natural log of 1 + 2 x plus c. Which is equal to 2 x + 1 divided by 2 multiplied by the natural log of 1 + 2X minus X + C. So now what we need to do for our 4th step is we need to evaluate from 0, so we're evaluating from 0 to 2, so 0 to 2, or specifically 0 to 1. So once again, at this point for step 4, we are evaluating from 0 to 1, and when we evaluate from 0 to 1, it will give us the integral evaluated from 0 to 1 of the natural log of 1 + 2 X. D X Fantastic. And this will be equal to. Drum roll parentheses 2 X. + 1, so 2 x + 1 divided by 2. Multiplied by the natural log of 1 + 2 x. And we need to. Add So actually rather we need to subtract X. And we need to note once again that we're evaluating from 0 to 1. Fantastic. And when we perform our evaluation, we will find that it's equal to parentheses of 3 divided by 2 multiplied by the natural log of 3 minus 1 minus 0, which will be equal to 3 divided by 2 multiplied by the natural log of 3 minus 1. So therefore we could write that the limit as n approaches infinity of the sum evaluated from k is equal to 1 ton of 1 divided by n multiplied by the natural log of 1 + 2 k divided by. N will be equal to 3 divided by 2 multiplied by the natural log of 3 minus 1, and this is our correct and final answer. Hooray, we did it. So going back to the top to review what we've learned, we could safely say that our final answer once again without a doubt will be. 3 divided by 2 multiplied by the natural log of 3 minus 1. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the limits in Exercise 9 and 10 by identifying them with definite integrals and evaluating the integrals.lim (n → ∞) Σ (from k=1 to n) ln √(1 + k/n)

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Evaluate the limits in Exercise 9 and 10 by identifying them with definite integrals and evaluating the integrals.
lim (n → ∞) Σ (from k=1 to n) ln √(1 + k/n)