![]() ![]() I'm not sure though, bc not I'm more confused after writing this comment.ĮDIT: Read my comment below! It has more examples!ĭivergence of a series does not always mean that the terms get larger or stay the same size. How I think about it is that 1/x^n, where n > 1, has the denominator of 1/k skipping more and more values (1, 1/4, 1/9,1/16, 1/25, 1/36, etc.) while 1/x passes through every value which sums up to infinity. This doesn't work with 1/x^2, for instance: Even if that didn't make any sense, think about this: the difference between any two consecutive values doesn't change very fast, so the sum is able to continually grow. ![]() The numbers are much larger, and they never seem to stop like convergent series do, even though it grows so slowly. Then, zoom out in the graph and go as far right as you'd like. It makes sense that the graph of ln(x) must approach infinity since you can plug any number into e^x (the "answer" for ln(x)). Have you ever looked at the graph of ln(x)? If you look on desmos ( ) and slowly scroll to the right, it feels like the values keep growing by smaller and smaller values, but nevertheless always growing. Basically, the terms in the sequence decrease towards 0 at a slower and slower rate, allowing the sum to approach infinity. My intuition for this is somewhat mentioned in the comment below. This is a little convoluted, the way I explained it, but I feel after reading this the concepts will become clearer Combine them so that the integral is in the middle, and you get: the integral < (P-Series < the integral +1. Now keep all those inequalities in mind, and add 1 to each to get (P-series) < the integral +1 < (P-series +1). So.if we took the P series minus 1 would be the right Riemann series from 1 to ∞ (smaller than ∫) and the standard P series would be the left Riemann series from 1 to ∞ (bigger than ∫) This statement is: (P-series -1)< the integral < (P-series). We do know that the first term of the P series will always be 1. But the problem is that it's not really a fair comparison after the shift since the P-series is now starting at 0 and going to ∞ instead of starting at 1 like the integral. Keep in mind the P series is equivalent to itself! We're just shifting the placement of it. But, the right Riemann can shift the P series over to the left by one. You can probably see why that is greater than the integral - there are the left over corners on the top. ![]() So, first, the left Riemann sum from 1 to ∞. The integral that we are working with is from 1 to ∞. And since the function is descending, it can be concluded that a left Riemann sum will be greater than a right Riemann sum. So, the key thing is that the P series, shown in Orange are essentially left and right Riemann sums. ![]()
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